package main

import (
	"bufio"
	"fmt"
	"os"
	"sort"
	"strconv"
)

var sc = bufio.NewScanner(os.Stdin)

func Scan() string {
	sc.Scan()
	return sc.Text()
}
func iScan() int {
	n, _ := strconv.Atoi(Scan())
	return n
}
func fScan() float64 {
	n, _ := strconv.ParseFloat(Scan(), 64)
	return n
}
func stringToInt(s string) int {
	n, _ := strconv.Atoi(s)
	return n
}
func SScan(n int) []string {
	a := make([]string, n)
	for i := 0; i < n; i++ {
		a[i] = Scan()
	}
	return a
}
func iSScan(n int) []int {
	a := make([]int, n)
	for i := 0; i < n; i++ {
		a[i] = iScan()
	}
	return a
}
func abs(x int) int {
	if x < 0 {
		x = -x
	}
	return x
}
func larger(a, b int) int {
	if a < b {
		return b
	} else {
		return a
	}
}
func smaller(a, b int) int {
	if a > b {
		return b
	} else {
		return a
	}
}
func max(a []int) int {
	x := a[0]
	for i := 0; i < len(a); i++ {
		if x < a[i] {
			x = a[i]
		}
	}
	return x
}
func min(a []int) int {
	x := a[0]
	for i := 0; i < len(a); i++ {
		if x > a[i] {
			x = a[i]
		}
	}
	return x
}
func sum(a []int) int {
	x := 0
	for _, v := range a {
		x += v
	}
	return x
}

var mod int = 1000000007

func main() {
	buf := make([]byte, 0)
	sc.Buffer(buf, mod)
	sc.Split(bufio.ScanWords)
	n := iScan()
	ab := make([][2]int, n+2)
	for i := 1; i <= n; i++ {
		ab[i] = [2]int{iScan(), iScan()}
	}
	ab[0] = [2]int{0, mod}
	ab[n+1] = [2]int{mod, -1}
	sort.Slice(ab, func(i, j int) bool { return ab[i][0] < ab[j][0] })
	fmt.Println(ab)
	ans := -1
	for i := 0; i < n+1; i++ {
		ans += smaller(ab[i+1][0]-ab[i][0], ab[i][1]-ab[i+1][1])
	}
	fmt.Println(ans)
}

Score : 200 points

Problem Statement

A group of people played a game. All players had distinct scores, which are positive integers.

Takahashi knows N facts on the players' scores. The i-th fact is as follows: the A_i-th highest score among the players is B_i.

Find the maximum possible number of players in the game.

Constraints

1 \leq N \leq 10^5

1 \leq A_i \leq 10^9(1\leq i\leq N)

0 \leq B_i \leq 10^9(1\leq i\leq N)

If i ≠ j, A_i ≠ A_j.

There exists a possible outcome of the game that are consistent with the facts.

All input values are integers.

Inputs

Input is given from Standard Input in the following format:

N
A_1 B_1
:
A_N B_N

Outputs

Print the maximum possible number of players in the game.

Sample Input 1

3
4 7
2 9
6 2

Sample Output 1

8

The maximum possible number of players is achieved when, for example, the players have the following scores: 12,9,8,7,5,2,1,0.

Sample Input 2

5
1 10
3 6
5 2
4 4
2 8

Sample Output 2

7

Sample Input 3

2
1 1000000000
1000000000 1

Sample Output 3

1000000001


package main

import (
	"bufio"
	"fmt"
	"os"
)

var sc = bufio.NewScanner(os.Stdin)

func ri() (n int) {
	sc.Scan()
	for _, v := range sc.Bytes() {
		n = n*10 + int(v-48)
	}
	return
}

func main() {
	sc.Split(bufio.ScanWords)
	n, k := ri(), ri()
	ors := make([]int, 0, 31)
	ors = append(ors, k)
	for i := 0; i <= 30; i++ {
		m := 1 << uint(i)
		if k&(m) != 0 {
			ors = append(ors, (k^m)|(m-1))
		}
	}
	sums := make([]int, len(ors))
	for i := 0; i < n; i++ {
		a, b := ri(), ri()
		for j, w := range ors {
			if w|a == w {
				sums[j] += b
			}
		}
	}
	max := sums[0]
	for _, v := range sums {
		if v > max {
			max = v
		}
	}
	fmt.Println(max)
}

Global OOD problem IDs are disjoint from train, validation, and in-distribution test problem IDs.

Score : 500 points

Problem Statement

Seisu-ya, a store specializing in non-negative integers, sells N non-negative integers. The i-th integer is A_i and has a utility of B_i.
There may be multiple equal integers with different utilities.

Takahashi will buy some integers in this store. He can buy a combination of integers whose bitwise OR is less than or equal to K. He wants the sum of utilities of purchased integers to be as large as possible.

Find the maximum possible sum of utilities of purchased integers.

Constraints

1 \leq N \leq 10^5

0 \leq K < 2^{30}

0 \leq A_i < 2^{30}(1\leq i\leq N)

1 \leq B_i \leq 10^9(1\leq i\leq N)

All input values are integers.

Inputs

Input is given from Standard Input in the following format:

N K
A_1 B_1
:
A_N B_N

Outputs

Print the maximum possible sum of utilities of purchased integers.

Sample Input 1

3 5
3 3
4 4
2 5

Sample Output 1

8

Buy 2 and 3 to achieve the maximum possible total utility, 8.

Sample Input 2

3 6
3 3
4 4
2 5

Sample Output 2

9

Buy 2 and 4 to achieve the maximum possible total utility, 9.

Sample Input 3

7 14
10 5
7 4
11 4
9 8
3 6
6 2
8 9

Sample Output 3

32

Submitted source-code file size, in bytes.

Execution CPU time reported by official Project CodeNet metadata, in milliseconds.

Peak memory usage reported by official Project CodeNet metadata, in kilobytes.


package main

import (
	"fmt"
	"sort"
)

type ins []in

type in struct {
	a int
	b int
	cp float64
}

func (ii ins) Len() int {
	return len(ii)
}

func (ii ins) Swap(i, j int) {
	ii[i], ii[j] = ii[j], ii[i]
}

func (ii ins) Less(i, j int) bool {
	if ii[i].cp == ii[j].cp {
		return ii[i].b > ii[j].b
	}
	return ii[i].cp > ii[j].cp
}

func main() {
	var n, k, ta, tb int
	fmt.Scan(&n, &k)
	s := make(ins, n)
	for key := range s {
		fmt.Scan(&s[key].a, &s[key].b)
		s[key].cp = float64(s[key].b) / float64(s[key].a)
	}
	sort.Sort(s)

	for _, value := range s {
		if ta == k {
			break
		}
		if ta | value.a <= k {
			ta = ta | value.a
			tb += value.b
		}
	}
	fmt.Print(tb)
}


// ProblemURL : https://atcoder.jp/contests/code-festival-2016-qualb/tasks/codefestival_2016_qualB_a
// ---------------------------------------------
package main

import "fmt"

func main() {
	var s string
	fmt.Scan(&s)
	if len(s) >= 4 && s[:4] == "YAKI" {
		fmt.Println("Yes")
	} else {
		fmt.Println("No")
	}
}

Score : 100 points

Problem Statement

Ringo is giving a present to Snuke.

Ringo has found out that Snuke loves yakiniku (a Japanese term meaning grilled meat. yaki: grilled, niku: meat). He supposes that Snuke likes grilled things starting with YAKI in Japanese, and does not like other things.

You are given a string S representing the Japanese name of Ringo's present to Snuke. Determine whether S starts with YAKI.

Constraints

1 \leq |S| \leq 10

S consists of uppercase English letters.

Input

Input is given from Standard Input in the following format:

S

Output

If S starts with YAKI, print Yes; otherwise, print No.

Sample Input 1

YAKINIKU

Sample Output 1

Yes

YAKINIKU starts with YAKI.

Sample Input 2

TAKOYAKI

Sample Output 2

No

TAKOYAKI (a Japanese snack. tako: octopus) does not start with YAKI.

Sample Input 3

YAK

Sample Output 3

No


package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
)

var sc = bufio.NewScanner(os.Stdin)

func scanInt() int {
	iv, _ := strconv.Atoi(scanString())
	return iv
}

func scanString() string {
	sc.Scan()
	return sc.Text()
}

func init() {
	sc.Split(bufio.ScanWords)
	sc.Buffer(make([]byte, 100001), 100001*100)
}

func main() {
	h, w := scanInt(), scanInt()
	xm := make(map[rune]int)
	for i := 0; i < h; i++ {
		a := scanString()
		for _, v := range a {
			xm[v]++
		}
	}

	var g1, g2, g4 int
	if h%2 == 1 && w%2 == 1 {
		g1 = 1
	}
	if h%2 == 1 {
		g2 += w / 2
	}
	if w%2 == 1 {
		g2 += h / 2
	}
	g4 = (h / 2) * (w / 2)

	for i := 0; i < g4; i++ {
		found := false
		for j := range xm {
			if xm[j] >= 4 {
				xm[j] -= 4
				found = true
				break
			}
		}
		if !found {
			fmt.Println("No")
			return
		}
	}
	for i := 0; i < g2; i++ {
		found := false
		for j := range xm {
			if xm[j] >= 2 {
				xm[j] -= 2
				found = true
				break
			}
		}
		if !found {
			fmt.Println("No")
			return
		}
	}
	if g1 == 1 {
		found := false
		for j := range xm {
			if xm[j] >= 1 {
				xm[j]--
				found = true
				break
			}
		}
		if !found {
			fmt.Println("No")
			return
		}
	}
	fmt.Println("Yes")
}

Score : 400 points

Problem Statement

We have an H-by-W matrix.
Let a_{ij} be the element at the i-th row from the top and j-th column from the left.
In this matrix, each a_{ij} is a lowercase English letter.

Snuke is creating another H-by-W matrix, A', by freely rearranging the elements in A.
Here, he wants to satisfy the following condition:

Every row and column in A' can be read as a palindrome.

Determine whether he can create a matrix satisfying the condition.

Note

A palindrome is a string that reads the same forward and backward.
For example, a, aa, abba and abcba are all palindromes, while ab, abab and abcda are not.

Constraints

1 ≤ H, W ≤ 100

a_{ij} is a lowercase English letter.

Input

Input is given from Standard Input in the following format:

H W
a_{11}a_{12}...a_{1W}
:
a_{H1}a_{H2}...a_{HW}

Output

If Snuke can create a matrix satisfying the condition, print Yes; otherwise, print No.

Sample Input 1

3 4
aabb
aabb
aacc

Sample Output 1

Yes

For example, the following matrix satisfies the condition.

abba
acca
abba

Sample Input 2

2 2
aa
bb

Sample Output 2

No

It is not possible to create a matrix satisfying the condition, no matter how we rearrange the elements in A.

Sample Input 3

5 1
t
w
e
e
t

Sample Output 3

Yes

For example, the following matrix satisfies the condition.

t
e
w
e
t

Sample Input 4

2 5
abxba
abyba

Sample Output 4

No

Sample Input 5

1 1
z

Sample Output 5

Yes


package main

import (
	"bufio"
	"fmt"
	"math"
	"os"
	"strconv"
)

//Util
func nextStr() string {
	sc.Scan()
	return sc.Text()
}

func nextInt() int {
	sc.Scan()
	i, e := strconv.Atoi(sc.Text())
	if e != nil {
		panic(e)
	}
	return i
}

func maxInt(a, b int) int {
	if a > b {
		return a
	} else {
		return b
	}
}

func minInt(a, b int) int {
	if a > b {
		return b
	} else {
		return a
	}
}

func absInt(a int) int {
	if a >= 0 {
		return a
	} else {
		return -a
	}
}

func absFloat64(a float64) float64 {
	if a >= 0 {
		return a
	} else {
		return -a
	}
}

func maxFloat64(a, b float64) float64 {
	if a > b {
		return a
	} else {
		return b
	}
}

func minFloat64(a, b float64) float64 {
	if a > b {
		return b
	} else {
		return a
	}
}

func str2Int(s string) int {
	n, _ := strconv.Atoi(s)
	return n
}

func reverse(s string) string {
	runes := []rune(s)
	for i, j := 0, len(runes)-1; i < j; i, j = i+1, j-1 {
		runes[i], runes[j] = runes[j], runes[i]
	}
	return string(runes)
}

func pow(p, q int) int {
	return int(math.Pow(float64(p), float64(q)))
}

func isPrime(x int) bool {
	if x == 1 {
		return false
	}
	if x == 2 {
		return true
	}
	if x%2 == 0 {
		return false
	}

	b := true
	rootx := int(math.Sqrt(float64(x)))
	i := 3
	for i <= rootx {
		if x%i == 0 {
			b = false
			break
		}
		i += 2
	}
	return b
}

//Main
const (
	initialBufSize = 10000
	maxBufSize     = 1000000
	mod            = 1e9 + 7
)

var (
	sc *bufio.Scanner = func() *bufio.Scanner {
		sc := bufio.NewScanner(os.Stdin)
		buf := make([]byte, initialBufSize)
		sc.Buffer(buf, maxBufSize)
		return sc
	}()
)

var N, K int

func main() {
	sc.Split(bufio.ScanWords)
	N, K = nextInt(), nextInt()

	ans := 0
	var tmp int
	for i := 0; i < N; i++ {
		tmp = nextInt()
		ans += 2 * minInt(K-tmp, tmp)
	}

	fmt.Println(ans)
}

Score : 200 points

Problem Statement

There are N balls in the xy-plane. The coordinates of the i-th of them is (x_i, i).
Thus, we have one ball on each of the N lines y = 1, y = 2, ..., y = N.

In order to collect these balls, Snuke prepared 2N robots, N of type A and N of type B.
Then, he placed the i-th type-A robot at coordinates (0, i), and the i-th type-B robot at coordinates (K, i).
Thus, now we have one type-A robot and one type-B robot on each of the N lines y = 1, y = 2, ..., y = N.

When activated, each type of robot will operate as follows.

When a type-A robot is activated at coordinates (0, a), it will move to the position of the ball on the line y = a, collect the ball, move back to its original position (0, a) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything.

When a type-B robot is activated at coordinates (K, b), it will move to the position of the ball on the line y = b, collect the ball, move back to its original position (K, b) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything.

Snuke will activate some of the 2N robots to collect all of the balls. Find the minimum possible total distance covered by robots.

Constraints

1 \leq N \leq 100

1 \leq K \leq 100

0 < x_i < K

All input values are integers.

Inputs

Input is given from Standard Input in the following format:

N
K
x_1 x_2 ... x_N

Outputs

Print the minimum possible total distance covered by robots.

Sample Input 1

1
10
2

Sample Output 1

4

There are just one ball, one type-A robot and one type-B robot.

If the type-A robot is used to collect the ball, the distance from the robot to the ball is 2, and the distance from the ball to the original position of the robot is also 2, for a total distance of 4.

Similarly, if the type-B robot is used, the total distance covered will be 16.

Thus, the total distance covered will be minimized when the type-A robot is used. The output should be 4.

Sample Input 2

2
9
3 6

Sample Output 2

12

The total distance covered will be minimized when the first ball is collected by the type-A robot, and the second ball by the type-B robot.

Sample Input 3

5
20
11 12 9 17 12

Sample Output 3

74


package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
)

const (
	initialBufSize = 100000
	maxBufSize     = 10000000
)

var (
	sc = bufio.NewScanner(os.Stdin)
	wt = bufio.NewWriter(os.Stdout)
)

func next() string {
	sc.Scan()
	return sc.Text()
}

func nextInt() int {
	i, _ := strconv.Atoi(next())
	return i
}

func nextInts(n int) []int {
	slice := make([]int, n)
	for i := 0; i < n; i++ {
		slice[i] = nextInt()
	}
	return slice
}

func puts(a ...interface{}) {
	fmt.Fprintln(wt, a...)
}

func main() {
	sc.Split(bufio.ScanWords)
	sc.Buffer(make([]byte, initialBufSize), maxBufSize)
	defer wt.Flush()

	const inf = 1 << 60

	n := nextInt()
	a := make([][]int, n)
	for i := range a {
		a[i] = nextInts(n)
		a[i][i] = inf
	}

	sum := 0
	for i := 0; i < n; i++ {
		for j := i + 1; j < n; j++ {
			// 経路i-k-j
			// 直通の道がなくても事足りるか
			eq := false
			for k := 0; k < n; k++ {
				if a[i][j] > a[i][k]+a[k][j] {
					puts(-1)
					return
				}
				eq = eq || (a[i][j] == a[i][k]+a[k][j])
			}
			if !eq {
				sum += a[i][j]
			}
		}
	}

	puts(sum)
}

Score : 500 points

Problem Statement

In Takahashi Kingdom, which once existed, there are N cities, and some pairs of cities are connected bidirectionally by roads.
The following are known about the road network:

People traveled between cities only through roads. It was possible to reach any city from any other city, via intermediate cities if necessary.

Different roads may have had different lengths, but all the lengths were positive integers.

Snuke the archeologist found a table with N rows and N columns, A, in the ruin of Takahashi Kingdom.
He thought that it represented the shortest distances between the cities along the roads in the kingdom.

Determine whether there exists a road network such that for each u and v, the integer A_{u, v} at the u-th row and v-th column of A is equal to the length of the shortest path from City u to City v.
If such a network exist, find the shortest possible total length of the roads.

Constraints

1 \leq N \leq 300

If i ≠ j, 1 \leq A_{i, j} = A_{j, i} \leq 10^9.

A_{i, i} = 0

Inputs

Input is given from Standard Input in the following format:

N
A_{1, 1} A_{1, 2} ... A_{1, N}
A_{2, 1} A_{2, 2} ... A_{2, N}
...
A_{N, 1} A_{N, 2} ... A_{N, N}

Outputs

If there exists no network that satisfies the condition, print -1.
If it exists, print the shortest possible total length of the roads.

Sample Input 1

3
0 1 3
1 0 2
3 2 0

Sample Output 1

3

The network below satisfies the condition:

City 1 and City 2 is connected by a road of length 1.

City 2 and City 3 is connected by a road of length 2.

City 3 and City 1 is not connected by a road.

Sample Input 2

3
0 1 3
1 0 1
3 1 0

Sample Output 2

-1

As there is a path of length 1 from City 1 to City 2 and City 2 to City 3, there is a path of length 2 from City 1 to City 3.
However, according to the table, the shortest distance between City 1 and City 3 must be 3.

Thus, we conclude that there exists no network that satisfies the condition.

Sample Input 3

5
0 21 18 11 28
21 0 13 10 26
18 13 0 23 13
11 10 23 0 17
28 26 13 17 0

Sample Output 3

82

Sample Input 4

3
0 1000000000 1000000000
1000000000 0 1000000000
1000000000 1000000000 0

Sample Output 4

3000000000


// Package main provides
//
// File:  d.go
// Author: ymiyamoto
//
// Created on Mon Jan  7 17:54:40 2019
//
package main

import (
	"errors"
	"fmt"
)

type graph [][]int

func warshalFloyd(g graph) (int, error) {
	g2 := make(graph, len(g))
	for i := range g2 {
		g2[i] = make([]int, len(g[i]))
		copy(g2[i], g[i])
	}

	for k := range g {
		for i := range g {
			for j := range g {
				if g[i][k]+g[k][j] < g[i][j] {
					return 0, errors.New("error")
				} else if i != k && k != j && g[i][k]+g[k][j] == g[i][j] {
					g2[i][j] = 0
				}
			}
		}
	}

	sum := 0
	for i := range g2 {
		for j := range g2[i] {
			sum += g2[i][j]
		}
	}

	return sum / 2, nil
}

func main() {
	var N int
	fmt.Scan(&N)

	routes := make(graph, N)
	for i := range routes {
		routes[i] = make([]int, N)
		for j := range routes[i] {
			fmt.Scan(&routes[i][j])
		}
	}

	if n, err := warshalFloyd(routes); err != nil {
		fmt.Println(-1)
	} else {
		fmt.Println(n)
	}
}


package main

import (
  "fmt"
  "strings"
)

func main() {
  var n string
  fmt.Scan(&n)
  i := strings.Index(n, "9")
  ans := "Yes"; if i == -1 {ans = "No"}
  fmt.Println(ans)

}

Score : 100 points

Problem Statement

It is September 9 in Japan now.

You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N?

Constraints

10≤N≤99

Input

Input is given from Standard Input in the following format:

N

Output

If 9 is contained in the decimal notation of N, print Yes; if not, print No.

Sample Input 1

29

Sample Output 1

Yes

The one's digit of 29 is 9.

Sample Input 2

72

Sample Output 2

No

72 does not contain 9.

Sample Input 3

91

Sample Output 3

Yes


package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
)

var sc = bufio.NewScanner(os.Stdin)

func Scan() string {
	sc.Scan()
	return sc.Text()
}
func iScan() int {
	n, _ := strconv.Atoi(Scan())
	return n
}
func fScan() float64 {
	n, _ := strconv.ParseFloat(Scan(), 64)
	return n
}
func stringToInt(s string) int {
	n, _ := strconv.Atoi(s)
	return n
}
func SScan(n int) []string {
	a := make([]string, n)
	for i := 0; i < n; i++ {
		a[i] = Scan()
	}
	return a
}
func iSScan(n int) []int {
	a := make([]int, n)
	for i := 0; i < n; i++ {
		a[i] = iScan()
	}
	return a
}
func abs(x int) int {
	if x < 0 {
		x = -x
	}
	return x
}
func larger(a, b int) int {
	if a < b {
		return b
	} else {
		return a
	}
}
func smaller(a, b int) int {
	if a > b {
		return b
	} else {
		return a
	}
}
func max(a []int) int {
	x := a[0]
	for i := 0; i < len(a); i++ {
		if x < a[i] {
			x = a[i]
		}
	}
	return x
}
func min(a []int) int {
	x := a[0]
	for i := 0; i < len(a); i++ {
		if x > a[i] {
			x = a[i]
		}
	}
	return x
}
func sum(a []int) int {
	x := 0
	for _, v := range a {
		x += v
	}
	return x
}

var mod int = 1000000007

func main() {
	buf := make([]byte, 0)
	sc.Buffer(buf, mod)
	sc.Split(bufio.ScanWords)
	n := iScan()
	d := make(map[int]struct{})
	for i := 0; i < n; i++ {
		a := iScan()
		if _, ok := d[a]; ok {
			delete(d, a)
		} else {
			d[a] = struct{}{}
		}
	}
	fmt.Println(len(d))
}

Score : 300 points

Problem Statement

You are playing the following game with Joisino.

Initially, you have a blank sheet of paper.

Joisino announces a number. If that number is written on the sheet, erase the number from the sheet; if not, write the number on the sheet. This process is repeated N times.

Then, you are asked a question: How many numbers are written on the sheet now?

The numbers announced by Joisino are given as A_1, ... ,A_N in the order she announces them. How many numbers will be written on the sheet at the end of the game?

Constraints

1≤N≤100000

1≤A_i≤1000000000(=10^9)

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
A_1
:
A_N

Output

Print how many numbers will be written on the sheet at the end of the game.

Sample Input 1

3
6
2
6

Sample Output 1

1

The game proceeds as follows:

6 is not written on the sheet, so write 6.

2 is not written on the sheet, so write 2.

6 is written on the sheet, so erase 6.

Thus, the sheet contains only 2 in the end. The answer is 1.

Sample Input 2

4
2
5
5
2

Sample Output 2

0

It is possible that no number is written on the sheet in the end.

Sample Input 3

6
12
22
16
22
18
12

Sample Output 3

2


package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
)

var sc = bufio.NewScanner(os.Stdin)

func nextInt() int {
	sc.Scan()
	i, _ := strconv.Atoi(sc.Text())
	return i
}

func inputIntSlice(size int) []int {
	slice := make([]int, size)
	for i := range slice {
		slice[i] = nextInt()
	}
	return slice
}

var written [1000000000]bool

func main() {
	sc.Split(bufio.ScanWords)
	n := nextInt()
	max := 0
	for i := 0; i < n; i++ {
		a := nextInt() - 1
		if a > max {
			max = a
		}
		written[a] = !written[a]
	}

	ans := 0
	for i := 0; i <= max; i++ {
		if written[i] {
			ans++
		}
	}
	fmt.Println(ans)
}


package main

import (
	"fmt"
)

func main() {
	var x, t int
	fmt.Scan(&x, &t)

	if v := x - t; v > 0 {
		fmt.Println(v)
	} else {
		fmt.Println(0)
	}
}

Score : 100 points

Problem Statement

We have a sandglass that runs for X seconds. The sand drops from the upper bulb at a rate of 1 gram per second. That is, the upper bulb initially contains X grams of sand.

How many grams of sand will the upper bulb contains after t seconds?

Constraints

1≤X≤10^9

1≤t≤10^9

X and t are integers.

Input

The input is given from Standard Input in the following format:

X t

Output

Print the number of sand in the upper bulb after t second.

Sample Input 1

100 17

Sample Output 1

83

17 out of the initial 100 grams of sand will be consumed, resulting in 83 grams.

Sample Input 2

48 58

Sample Output 2

0

All 48 grams of sand will be gone, resulting in 0 grams.

Sample Input 3

1000000000 1000000000

Sample Output 3

0


package main

import (
	"bufio"
	"container/heap"
	"fmt"
	"os"
	"sort"
	"strconv"
)

const (
	initialBufSize = 1e4
	maxBufSize     = 1e6 + 7
)

var buf []byte = make([]byte, initialBufSize)
var sc = bufio.NewScanner(os.Stdin)

func init() {
	sc.Split(bufio.ScanWords)
	sc.Buffer(buf, maxBufSize)
}

var d = []Position{{1, 0}, {-1, 0}, {0, 1}, {0, -1}}
var d8 = []Position{{1, 0}, {-1, 0}, {0, 1}, {0, -1}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}}

func main() {
	N := nextInt()
	var cnt, serial int
	serial++
	var prev bool
	for i := 0; i < N; i++ {
		p := nextInt() - 1
		if i == p {
			cnt++
			if prev {
				serial++
			} else {
				prev = true
			}
		}
		if i != p || (i == N-1 && prev) {
			prev = false
			cnt -= serial / 2
			serial = 1
		}
	}
	fmt.Println(cnt)
}

// Position として迷路問題での現在地を表す構造体を定義
type Position struct {
	H int
	W int
}

func nextLine() string {
	sc.Scan()
	return sc.Text()
}

func nextInt() int {
	sc.Scan()
	i, e := strconv.Atoi(sc.Text())
	if e != nil {
		panic(e)
	}
	return i
}

func abs(a int) int {
	if a > 0 {
		return a
	} else {
		return -a
	}
}

func pow(p, q int) int {
	res := p
	if q == 0 {
		return 1
	}
	for i := 0; i < q-1; i++ {
		res *= p
	}
	return res
}

func ceil(a, b int) int {
	return (a + (b - 1)) / b
}

func MinOf(vars ...int) int {
	min := vars[0]
	for _, i := range vars {
		if min > i {
			min = i
		}
	}
	return min
}

func MaxOf(vars ...int) int {
	max := vars[0]
	for _, i := range vars {
		if max < i {
			max = i
		}
	}
	return max
}

func gcdof2numbers(a int, b int) int {
	if b == 0 {
		return a
	}
	return gcdof2numbers(b, a%b)
}

func lcmof2numbers(a int, b int) int {
	return a / gcdof2numbers(a, b) * b
}

// マイナスの場合は考慮していない
func fuctorial(a int) int {
	if a == 1 || a == 0 {
		return 1
	} else {
		return fuctorial(a-1) * a
	}
}

// NextPermutation はsortインターフェース（要はソート済みスライス）から次の順列にスライスを並べ替える
// https://play.golang.org/p/ljft9xhOEn
func NextPermutation(x sort.Interface) bool {
	n := x.Len() - 1
	if n < 1 {
		return false
	}
	j := n - 1
	for ; !x.Less(j, j+1); j-- {
		if j == 0 {
			return false
		}
	}
	l := n
	for !x.Less(j, l) {
		l--
	}
	x.Swap(j, l)
	for k, l := j+1, n; k < l; {
		x.Swap(k, l)
		k++
		l--
	}
	return true
}

// PrimeFactorize 素因数分解したmapを返す
func PrimeFactorize(n int) map[int]int {
	pf := map[int]int{}
	for i := 2; i*i <= n; i++ {
		for n%i == 0 {
			pf[i]++
			n /= i
		}
	}
	if n != 1 {
		pf[n]++
	}
	return pf
}

// FastPrimeFactorize osa_k法を使った高速素因数分解
// sieveを作るところのjはi*2スタートでもよい？
// https://qiita.com/rsk0315_h4x/items/ff3b542a4468679fb409
func FastPrimeFactorize(n int, sieve []int) map[int]int {
	pf := make(map[int]int)
	for n > 1 {
		pf[sieve[n]]++
		n /= sieve[n]
	}
	return pf
}

// MakeSieve は高速素数判定に使う篩を作る。
// Sieve[i] < i なら合成数なので、Sieveは素数判定にも使える
func MakeSieve(max int) []int {
	sieve := make([]int, max)
	for i := range sieve {
		sieve[i] = i
	}
	for i := 2; i*i <= max; i++ {
		if sieve[i] == i {
			for j := i; j < max; j += i {
				if sieve[j] == j {
					sieve[j] = i
				}
			}
		}
	}
	return sieve
}

// golangの公式サンプルより
// https://xn--go-hh0g6u.com/pkg/container/heap/#example__intHeap
// IntHeap は，整数の最小ヒープです。
type IH []int

func (h IH) Len() int           { return len(h) }
func (h IH) Less(i, j int) bool { return h[i] < h[j] }
func (h IH) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }

func (h *IH) Push(x interface{}) {
	// Push と Pop はポインタレシーバを使っています。
	// なぜなら，スライスの中身だけでなく，スライスの長さも変更するからです。
	*h = append(*h, x.(int))
}

func (h *IH) Pop() interface{} {
	old := *h
	n := len(old)
	x := old[n-1]
	*h = old[0 : n-1]
	return x
}

// golangの公式サンプルより
// https://xn--go-hh0g6u.com/pkg/container/heap/#example__priorityQueue

// Item は，優先キューで管理する項目です。
type Item struct {
	value    string // 値。任意です。
	priority int    // キューにおける優先度
	// index は heap.Interface メソッドで更新されます。
	index int // ヒープにおけるインデックス
}

// PriorityQueue は heap.Interface を実装し，Item のリストを保持します。
type PriorityQueue []*Item

func (pq PriorityQueue) Len() int { return len(pq) }

func (pq PriorityQueue) Less(i, j int) bool {
	// Pop が最小ではなく最大の優先度を持つ項目を返して欲しいので，ここでは > を使っています。
	return pq[i].priority > pq[j].priority
}

func (pq PriorityQueue) Swap(i, j int) {
	pq[i], pq[j] = pq[j], pq[i]
	pq[i].index = i
	pq[j].index = j
}

func (pq *PriorityQueue) Push(x interface{}) {
	n := len(*pq)
	item := x.(*Item)
	item.index = n
	*pq = append(*pq, item)
}

func (pq *PriorityQueue) Pop() interface{} {
	old := *pq
	n := len(old)
	item := old[n-1]
	old[n-1] = nil  // メモリリークを避ける
	item.index = -1 // 安全のため
	*pq = old[0 : n-1]
	return item
}

// update はキューの Item の優先度と値を更新します。
func (pq *PriorityQueue) update(item *Item, value string, priority int) {
	item.value = value
	item.priority = priority
	heap.Fix(pq, item.index)
}

// UnionFind の定義
// 下記を参考
// https://youtu.be/TdR816rqc3s?t=6822
// https://github.com/atcoder/live_library/blob/master/uf.cpp
// https://qiita.com/haru1843/items/2295d0ec1f5002bd5c33#%E5%AE%9F%E8%A3%85
type UnionFind struct {
	parent []int
	maxlen int
}

// UnionFind のスライス初期化
func newUnionFind(N int) *UnionFind {
	u := new(UnionFind)
	u.parent = make([]int, N)
	u.maxlen = 1
	for i := range u.parent {
		u.parent[i] = -1
	}
	return u
}

// xの根を見つける
func (u *UnionFind) find(x int) int {
	if u.parent[x] < 0 {
		return x
	}
	u.parent[x] = u.find(u.parent[x])
	return u.parent[x]
}

// xとyのグループを結合する
func (u *UnionFind) unite(x, y int) {
	xf := u.find(x)
	yf := u.find(y)
	if xf == yf {
		return
	}
	// 常に大きい方に小さい木をくっつける
	if xf > yf {
		xf, yf = yf, xf
	}
	u.parent[xf] += u.parent[yf]
	u.parent[yf] = xf
	if u.parent[xf] < 0 {
		u.maxlen = MaxOf(u.maxlen, abs(u.parent[xf]))
	}
}

// xとyが同じグループに所属するかどうかを返す
func (u *UnionFind) same(x, y int) bool {
	if u.find(x) == u.find(y) {
		return true
	}
	return false
}

// xの所属するグループの木の大きさを返す
func (u *UnionFind) size(x int) int {
	return -u.parent[u.find(x)]
}

// 根の数を返す
func (u UnionFind) rootcnt() int {
	var cnt int
	for _, v := range u.parent {
		if v < 0 {
			cnt++
		}
	}
	return cnt
}

Score : 400 points

Problem Statement

You are given a permutation p_1,p_2,...,p_N consisting of 1,2,..,N.
You can perform the following operation any number of times (possibly zero):

Operation: Swap two adjacent elements in the permutation.

You want to have p_i ≠ i for all 1≤i≤N.
Find the minimum required number of operations to achieve this.

Constraints

2≤N≤10^5

p_1,p_2,..,p_N is a permutation of 1,2,..,N.

Input

The input is given from Standard Input in the following format:

N
p_1 p_2 .. p_N

Output

Print the minimum required number of operations

Sample Input 1

5
1 4 3 5 2

Sample Output 1

2

Swap 1 and 4, then swap 1 and 3. p is now 4,3,1,5,2 and satisfies the condition.
This is the minimum possible number, so the answer is 2.

Sample Input 2

2
1 2

Sample Output 2

1

Swapping 1 and 2 satisfies the condition.

Sample Input 3

2
2 1

Sample Output 3

0

The condition is already satisfied initially.

Sample Input 4

9
1 2 4 9 5 8 7 3 6

Sample Output 4

3


package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
)

func min(nums ...int) int {
	ret := nums[0]
	for _, v := range nums {
		if v < ret {
			ret = v
		}
	}
	return ret
}

func main() {
	sc.Split(bufio.ScanWords)
	q, h, s, d := nextInt(), nextInt(), nextInt(), nextInt()
	n := nextInt()
	puts(min(n*4*q, n*2*h, n*s, (n/2)*d+(n%2)*min(4*q, 2*h, s)))
	wt.Flush()
}

var (
	sc  = bufio.NewScanner(os.Stdin)
	rdr = bufio.NewReaderSize(os.Stdin, 1000000)
	wt  = bufio.NewWriter(os.Stdout)
)

func next() string {
	sc.Scan()
	return sc.Text()
}

func nextInt() int {
	i, _ := strconv.Atoi(next())
	return i
}

func nextFloat64() float64 {
	f, _ := strconv.ParseFloat(next(), 64)
	return f
}

func nextInts(n int) []int {
	slice := make([]int, n)
	for i := 0; i < n; i++ {
		slice[i] = nextInt()
	}
	return slice
}

func nextFloat64s(n int) []float64 {
	slice := make([]float64, n)
	for i := 0; i < n; i++ {
		slice[i] = nextFloat64()
	}
	return slice
}

func nextMega() string {
	buf := make([]byte, 0, 1000000)
	for {
		l, p, _ := rdr.ReadLine()
		buf = append(buf, l...)
		if !p {
			break
		}
	}
	return string(buf)
}

func putf(format string, a ...interface{}) {
	fmt.Fprintf(wt, format, a...)
}

func puts(a ...interface{}) {
	fmt.Fprintln(wt, a...)
}

Score : 300 points

Problem Statement

You've come to your favorite store Infinitesco to buy some ice tea.

The store sells ice tea in bottles of different volumes at different costs.
Specifically, a 0.25-liter bottle costs Q yen, a 0.5-liter bottle costs H yen, a 1-liter bottle costs S yen, and a 2-liter bottle costs D yen.
The store has an infinite supply of bottles of each type.

You want to buy exactly N liters of ice tea. How many yen do you have to spend?

Constraints

1 \leq Q, H, S, D \leq 10^8

1 \leq N \leq 10^9

All input values are integers.

Input

Input is given from Standard Input in the following format:

Q H S D
N

Output

Print the smallest number of yen you have to spend to buy exactly N liters of ice tea.

Sample Input 1

20 30 70 90
3

Sample Output 1

150

Buy one 2-liter bottle and two 0.5-liter bottles. You'll get 3 liters for 90 + 30 + 30 = 150 yen.

Sample Input 2

10000 1000 100 10
1

Sample Output 2

100

Even though a 2-liter bottle costs just 10 yen, you need only 1 liter.
Thus, you have to buy a 1-liter bottle for 100 yen.

Sample Input 3

10 100 1000 10000
1

Sample Output 3

40

Now it's better to buy four 0.25-liter bottles for 10 + 10 + 10 + 10 = 40 yen.

Sample Input 4

12345678 87654321 12345678 87654321
123456789

Sample Output 4

1524157763907942


package main

import (
	"bufio"
	"fmt"
	"os"
	"sort"
	"strconv"
)

func main() {
	var N int
	fmt.Scan(&N)
	set := make(map[int]int, 0)
	sc := bufio.NewScanner(os.Stdin)
	sc.Split(bufio.ScanWords)
	for i := 0; i < N; i++ {
		sc.Scan()
		t, _ := strconv.Atoi(sc.Text())
		set[t]++
	}
	ks := make([]int, 0)
	for k := range set {
		ks = append(ks, k)
	}
	sort.Sort(sort.Reverse(sort.IntSlice(ks)))
	m1, m2 := 0, 0
	for i := range ks {
		v := set[ks[i]]
		if v >= 2 {
			if v > m1 {
				m1 = ks[i]
			} else if v > m2 {
				m2 = ks[i]
			}
		}
	}
	ans := 0
	if set[m1] >= 4 {
		ans = m1 * m1
	} else {
		ans = m1 * m2
	}
	fmt.Println(ans)
}

Score : 300 points

Problem Statement

We have N sticks with negligible thickness.
The length of the i-th stick is A_i.

Snuke wants to select four different sticks from these sticks and form a rectangle (including a square), using the sticks as its sides.
Find the maximum possible area of the rectangle.

Constraints

4 \leq N \leq 10^5

1 \leq A_i \leq 10^9

A_i is an integer.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the maximum possible area of the rectangle.
If no rectangle can be formed, print 0.

Sample Input 1

6
3 1 2 4 2 1

Sample Output 1

2

1 \times 2 rectangle can be formed.

Sample Input 2

4
1 2 3 4

Sample Output 2

0

No rectangle can be formed.

Sample Input 3

10
3 3 3 3 4 4 4 5 5 5

Sample Output 3

20


package main

import (
	"fmt"
	"strconv"
)

func main() {
	var n int
	fmt.Scan(&n)

	s := strconv.Itoa(n)

	for i := 0; i < len(s)/2; i++ {
		if s[i] != s[len(s)-1-i] {
			fmt.Println("No")
			return
		}

	}
	fmt.Println("Yes")
}

Score : 100 points

Problem Statement

You are given a three-digit positive integer N.

Determine whether N is a palindromic number.

Here, a palindromic number is an integer that reads the same backward as forward in decimal notation.

Constraints

100≤N≤999

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

If N is a palindromic number, print Yes; otherwise, print No.

Sample Input 1

575

Sample Output 1

Yes

N=575 is also 575 when read backward, so it is a palindromic number. You should print Yes.

Sample Input 2

123

Sample Output 2

No

N=123 becomes 321 when read backward, so it is not a palindromic number. You should print No.

Sample Input 3

812

Sample Output 3

No


package main

import "fmt"

func main() {
	var n string
	fmt.Scan(&n)

	if n[0] == n[2] {
		fmt.Println("Yes")
	} else {
		fmt.Println("No")
	}
}


package main

import (
	"fmt"
)

func main() {
	var n string
	fmt.Scan(&n)

	if n[0] == n[2] {
		fmt.Println("Yes")
	} else {
		fmt.Println("No")
	}
}


// ProblemURL : https://atcoder.jp/contests/abc070/tasks/abc070_d
// ---------------------------------------------
package main

import (
	"bufio"
	"fmt"
	"math"
	"os"
	"sort"
	"strconv"
	"strings"
)

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}
func diff(a, b int) int {
	if a > b {
		return a - b
	}
	return b - a
}
func larger(a, b int) int {
	if a > b {
		return a
	}
	return b
}
func smaller(a, b int) int {
	if a < b {
		return a
	}
	return b
}
func largest(a, b, c int) (lgst int) {
	if a > b {
		lgst = a
	} else {
		lgst = b
	}
	if c > lgst {
		lgst = c
	}
	return
}
func smallest(a, b, c int) (slst int) {
	if a < b {
		slst = a
	} else {
		slst = b
	}
	if c < slst {
		slst = c
	}
	return
}
func intsMax(a []int) (val int) {
	if len(a) == 0 {
		panic("func intsMax: argument slice length must not be zero")
	}
	val = a[0]
	for _, aa := range a {
		if aa > val {
			val = aa
		}
	}
	return
}
func intsMaxIdx(a []int) (idx, val int) {
	if len(a) == 0 {
		panic("func intsMaxIdx: argument slice length must not be zero")
	}
	val = a[0]
	for i, aa := range a {
		if aa > val {
			idx, val = i, aa
		}
	}
	return
}
func intsMin(a []int) (val int) {
	if len(a) == 0 {
		panic("func intsMin: argument slice length must not be zero")
	}
	val = a[0]
	for _, aa := range a {
		if aa < val {
			val = aa
		}
	}
	return
}
func intsMinIdx(a []int) (idx, val int) {
	if len(a) == 0 {
		panic("func intsMinIdx: argument slice length must not be zero")
	}
	val = a[0]
	for i, aa := range a {
		if aa < val {
			idx, val = i, aa
		}
	}
	return
}
func intsSum(a []int) int {
	res := 0
	for _, v := range a {
		res += v
	}
	return res
}
func intsAve(s []int) float64 {
	return float64(intsSum(s)) / float64(len(s))
}
func intsAdd(a *[]int, x int)    { *a = append(*a, x) }
func intsDelete(a *[]int, i int) { *a = append((*a)[:i], (*a)[i+1:]...) }
func intsJoin(a, b []int) []int  { return append(a, b...) }
func intsClear(a []int) []int    { return a[:0] }
func intsCopy(a []int) []int     { return append([]int(nil), a...) }
func intsCopySorted(a []int) []int {
	s := intsCopy(a)
	sort.Ints(s)
	return s
}
func intsReverse(a []int) {
	for i, j := 0, len(a)-1; i < j; i, j = i+1, j-1 {
		a[i], a[j] = a[j], a[i]
	}
	return
}
func intsFill(a []int, val int) {
	for i := 0; i < len(a); i++ {
		a[i] = val
	}
	return
}
func intsPeekBack(a []int) int {
	if len(a) == 0 {
		panic("func peekBack: zero length slice")
	}
	return a[len(a)-1]
}
func intsPeekFront(a []int) int {
	if len(a) == 0 {
		panic("func peekFront: zero length slice")
	}
	return a[0]
}
func intsPopBack(a *[]int) int {
	if len(*a) == 0 {
		panic("func popBack: zero length slice")
	}
	t := (*a)[len(*a)-1]
	*a = (*a)[:len(*a)-1]
	return t
}
func intsPopFront(a *[]int) int {
	if len(*a) == 0 {
		panic("func popFront: zero length slice")
	}
	h := (*a)[0]
	*a = (*a)[1:]
	return h
}
func intsPushBack(a *[]int, x int) { *a = append(*a, x) }
func intsPushFront(a *[]int, x int) {
	*a = append(*a, 0)
	copy((*a)[1:], (*a)[0:])
	(*a)[0] = x
}
func intsAppendSentinelHead(a *[]int, sentinel int) { intsPushFront(a, sentinel) }
func intsAppendSentinelTail(a *[]int, sentinel int) { intsPushBack(a, sentinel) }
func intsAppendSentinel(a *[]int, sentinel int) {
	intsPushFront(a, sentinel)
	intsPushBack(a, sentinel)
}
func isEven(n int) bool    { return n&1 == 0 }
func isOdd(n int) bool     { return n&1 == 1 }
func floor(x float64) int  { return int(x) }
func ceil(x float64) int   { return int(math.Ceil(x)) }
func ceilDiv(x, y int) int { return (x + y - 1) / y }
func pow(x, y int) int {
	if y < 0 {
		panic("Exponent must be a natural number")
	}
	z := 1
	for i := y; i != 0; i >>= 1 {
		if i&1 == 1 {
			z *= x
		}
		x *= x
	}
	return z
}
func fact(x int) int {
	fact := 1
	for i := 1; i <= x; i++ {
		fact *= i
	}
	return fact
}
func sigmaK(n int) int  { return n * (n + 1) / 2 }
func sigmaKK(n int) int { return (n * (n + 1) / 2) * (2*n + 1) / 3 }
func euclideanDistance(a, b point) float64 {
	s := diff(a.x, b.x)
	s *= s
	t := diff(a.y, b.y)
	t *= t
	return math.Sqrt(float64(s) + float64(t))
}
func manhattanDistance(a, b point) int {
	s := diff(a.x, b.x)
	t := diff(a.y, b.y)
	return s + t
}
func swap(a, b *int) { *a, *b = *b, *a }
func chmax(a *int, b int) {
	if *a < b {
		*a = b
	}
}
func chmin(a *int, b int) {
	if *a > b {
		*a = b
	}
}
func cumSum(a []int) []int {
	b := append([]int{0}, a...)
	for i := 1; i < len(b); i++ {
		b[i] += b[i-1]
	}
	return b
}
func runesReverse(a []rune) {
	for i, j := 0, len(a)-1; i < j; i, j = i+1, j-1 {
		a[i], a[j] = a[j], a[i]
	}
	return
}
func bytesReverse(a []byte) {
	for i, j := 0, len(a)-1; i < j; i, j = i+1, j-1 {
		a[i], a[j] = a[j], a[i]
	}
	return
}
func strUpperCase(s string) string { return strings.ToUpper(s) }
func strLowerCase(s string) string { return strings.ToLower(s) }
func strToInt(s string) int {
	i, err := strconv.Atoi(s)
	if err != nil {
		panic(err)
	}
	return i
}
func intToStr(i int) string  { return strconv.Itoa(i) }
func CalcMod(a, mod int) int { return (a%mod + mod) % mod }
func ModPow(x, y, mod int) int {
	if y < 0 {
		panic("Exponent must be a natural number")
	}
	z := 1
	for i := y; i != 0; i >>= 1 {
		if i&1 == 1 {
			z = (z * x) % mod
		}
		x = (x * x) % mod
	}
	return z
}
func ModFact(x, mod int) int {
	fact := 1
	for i := 1; i <= x; i++ {
		fact = fact * i % mod
	}
	return fact
}
func ModInv(x, mod int) int { return ModPow(x, mod-2, mod) }
func initMatrixBool(parentSize, childSize int, initialValue bool) *[][]bool {
	res := make([][]bool, parentSize)
	for i := range res {
		res[i] = make([]bool, childSize)
		for j := range res[i] {
			res[i][j] = initialValue
		}
	}
	return &res
}
func initMatrixInt(parentSize, childSize int, initialValue int) *[][]int {
	res := make([][]int, parentSize)
	for i := range res {
		res[i] = make([]int, childSize)
		for j := range res[i] {
			res[i][j] = initialValue
		}
	}
	return &res
}

var (
	sc = bufio.NewScanner(os.Stdin)
	bw = bufio.NewWriterSize(os.Stdout, bwBufSize)
)

func init() {
	sc.Split(bufio.ScanWords)
	sc.Buffer(make([]byte, 1e7), 1e7)
}
func ru() (n int) {
	sc.Scan()
	if err := sc.Err(); err != nil {
		panic(err)
	}
	for _, v := range sc.Bytes() {
		n = n*10 + int(v-48)
	}
	return
}
func ri() (n int) {
	sc.Scan()
	if err := sc.Err(); err != nil {
		panic(err)
	}
	b := sc.Bytes()
	neg := false
	if b[0] == 45 {
		neg = true
		b = b[1:]
	}
	for _, v := range b {
		n = n*10 + int(v-48)
	}
	if neg {
		n = -n
	}
	return
}
func ri64() (n int64) {
	sc.Scan()
	if err := sc.Err(); err != nil {
		panic(err)
	}
	b := sc.Bytes()
	neg := false
	if b[0] == 45 {
		neg = true
		b = b[1:]
	}
	for _, v := range b {
		n = n*10 + int64(v-48)
	}
	if neg {
		n = -n
	}
	return
}
func rf64() float64 {
	sc.Scan()
	if err := sc.Err(); err != nil {
		panic(err)
	}
	f, err := strconv.ParseFloat(sc.Text(), 64)
	if err != nil {
		panic(err)
	}
	return f
}
func rs() string {
	sc.Scan()
	if err := sc.Err(); err != nil {
		panic(err)
	}
	return sc.Text()
}
func rb() []byte {
	sc.Scan()
	if err := sc.Err(); err != nil {
		panic(err)
	}
	return sc.Bytes()
}
func rr() []rune {
	sc.Scan()
	if err := sc.Err(); err != nil {
		panic(err)
	}
	return []rune(sc.Text())
}
func ris(n int) []int {
	s := make([]int, n)
	for i := range s {
		s[i] = ri()
	}
	return s
}
func risZeroBased(n int) []int {
	s := make([]int, n)
	for i := range s {
		s[i] = ri() - 1
	}
	return s
}
func ri64s(n int) []int64 {
	s := make([]int64, n)
	for i := range s {
		s[i] = ri64()
	}
	return s
}
func rss(n int) []string {
	s := make([]string, n)
	for i := range s {
		s[i] = rs()
	}
	return s
}
func pf(format string, a ...interface{}) {
	if _, err := fmt.Fprintf(bw, format, a...); err != nil {
		panic(err)
	}
}
func pln(a ...interface{}) {
	if _, err := fmt.Fprintln(bw, a...); err != nil {
		panic(err)
	}
}
func pints(a []int) {
	for _, v := range a {
		fmt.Fprintln(bw, v)
	}
}
func pintsol(a []int) {
	s := fmt.Sprint(a)
	if _, err := fmt.Fprintln(bw, s[1:len(s)-1]); err != nil {
		panic(err)
	}
}
func dbg(a ...interface{}) {
	if _, err := fmt.Fprintln(os.Stderr, a...); err != nil {
		panic(err)
	}
}
func dbgMatrix(a [][]int) {
	for i := 0; i < len(a); i++ {
		dbg(a[i])
	}
}
func YESorNO(b bool) string {
	if b {
		return "YES"
	} else {
		return "NO"
	}
}
func YesOrNo(b bool) string {
	if b {
		return "Yes"
	} else {
		return "No"
	}
}
func main() {
	solve()
	bw.Flush()
}

const (
	bwBufSize = 1e6

	alphabetLower = "abcdefghijklmnopqrstuvwxyz"
	alphabetUpper = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"

	// maxInt64  = 9223372036854775807  // 9e18
	// maxInt32  = 2147483647           // 2e9
	// maxUint64 = 18446744073709551615 // 1e19
	inf = 1 << 60
	mod = 1e9 + 7
)

type pair struct{ fi, se int }
type byFiSe []pair

func (p byFiSe) Len() int      { return len(p) }
func (p byFiSe) Swap(i, j int) { p[i], p[j] = p[j], p[i] }
func (p byFiSe) Less(i, j int) bool {
	if p[i].fi == p[j].fi {
		return p[i].se < p[j].se
	}
	return p[i].fi < p[j].fi
}

type point struct{ x, y int }

func solve() {
	n := ri()
	type edge struct{ t, c int }
	g := make([][]edge, n)
	for i := 0; i < n-1; i++ {
		a, b, c := ri()-1, ri()-1, ri()
		g[a] = append(g[a], edge{t: b, c: c})
		g[b] = append(g[b], edge{t: a, c: c})
	}
	q, k := ri(), ri()-1
	a := make([]int, n)
	var dfs func(int, int, int)
	dfs = func(v, p, d int) {
		a[v] = d
		for _, nv := range g[v] {
			if nv.t == p {
				continue
			}
			dfs(nv.t, v, d+nv.c)
		}
	}
	dfs(k, -1, 0)
	for i := 0; i < q; i++ {
		x, y := ri()-1, ri()-1
		pln(a[x] + a[y])
	}
}

Score : 400 points

Problem Statement

You are given a tree with N vertices.

Here, a tree is a kind of graph, and more specifically, a connected undirected graph with N-1 edges, where N is the number of its vertices.

The i-th edge (1≤i≤N-1) connects Vertices a_i and b_i, and has a length of c_i.

You are also given Q queries and an integer K. In the j-th query (1≤j≤Q):

find the length of the shortest path from Vertex x_j and Vertex y_j via Vertex K.

Constraints

3≤N≤10^5

1≤a_i,b_i≤N (1≤i≤N-1)

1≤c_i≤10^9 (1≤i≤N-1)

The given graph is a tree.

1≤Q≤10^5

1≤K≤N

1≤x_j,y_j≤N (1≤j≤Q)

x_j≠y_j (1≤j≤Q)

x_j≠K,y_j≠K (1≤j≤Q)

Input

Input is given from Standard Input in the following format:

N
a_1 b_1 c_1
:
a_{N-1} b_{N-1} c_{N-1}
Q K
x_1 y_1
:
x_{Q} y_{Q}

Output

Print the responses to the queries in Q lines.

In the j-th line j(1≤j≤Q), print the response to the j-th query.

Sample Input 1

5
1 2 1
1 3 1
2 4 1
3 5 1
3 1
2 4
2 3
4 5

Sample Output 1

3
2
4

The shortest paths for the three queries are as follows:

Query 1: Vertex 2 → Vertex 1 → Vertex 2 → Vertex 4 : Length 1+1+1=3

Query 2: Vertex 2 → Vertex 1 → Vertex 3 : Length 1+1=2

Query 3: Vertex 4 → Vertex 2 → Vertex 1 → Vertex 3 → Vertex 5 : Length 1+1+1+1=4

Sample Input 2

7
1 2 1
1 3 3
1 4 5
1 5 7
1 6 9
1 7 11
3 2
1 3
4 5
6 7

Sample Output 2

5
14
22

The path for each query must pass Vertex K=2.

Sample Input 3

10
1 2 1000000000
2 3 1000000000
3 4 1000000000
4 5 1000000000
5 6 1000000000
6 7 1000000000
7 8 1000000000
8 9 1000000000
9 10 1000000000
1 1
9 10

Sample Output 3

17000000000


package main

import (
	"bufio"
	"fmt"
	"io"
	"os"
	"strconv"
)

var (
	readString func() string
	stdout     *bufio.Writer
)

func init() {
	readString = newReadString(os.Stdin)
	stdout = bufio.NewWriter(os.Stdout)
}

func newReadString(ior io.Reader) func() string {
	r := bufio.NewScanner(ior)
	r.Buffer(make([]byte, 1024), int(1e+11))
	r.Split(bufio.ScanWords)

	return func() string {
		if !r.Scan() {
			panic("Scan failed")
		}
		return r.Text()
	}
}

func readInt() int {
	return int(readInt64())
}

func readInt64() int64 {
	i, err := strconv.ParseInt(readString(), 0, 64)
	if err != nil {
		panic(err.Error())
	}
	return i
}

func readIntSlice(n int) []int {
	b := make([]int, n)
	for i := 0; i < n; i++ {
		b[i] = readInt()
	}
	return b
}

func readLengthAndSlice() (int, []int) {
	n := readInt()
	return n, readIntSlice(n)
}

func printf(f string, args ...interface{}) (int, error) {
	return fmt.Fprintf(stdout, f, args...)
}

func println(args ...interface{}) (int, error) {
	return fmt.Fprintln(stdout, args...)
}

func eprintln(args ...interface{}) (int, error) {
	return fmt.Fprintln(os.Stderr, args...)
}

// readString() string
// readInt() int
// readIntSlice(n int) []int
// readLengthAndSlice() []int

// -----------------------------------------------------------------------------

func main() {
	defer stdout.Flush()
	_, a := readLengthAndSlice()

	c := make([]int, 4)
	for i := range a {
		c[a[i]%4]++
	}
	c1 := c[1] + c[3]
	c2 := c[2]
	c4 := c[0]
	if c2 == 0 {
		if c1 <= c4+1 {
			println("Yes")
		} else {
			println("No")
		}
	} else {
		if c2 == 1 && c4 == 0 {
			println("No")
		} else if c1 <= c4 {
			println("Yes")
		} else {
			println("No")
		}
	}
}

// -----------------------------------------------------------------------------

func minInt(a, b int) int {
	if a < b {
		return a
	}
	return b
}

func minUint(a, b uint) uint {
	if a < b {
		return a
	}
	return b
}

func minInt64(a, b int64) int64 {
	if a < b {
		return a
	}
	return b
}

func minUint64(a, b uint64) uint64 {
	if a < b {
		return a
	}
	return b
}

func maxInt(a, b int) int {
	if a < b {
		return b
	}
	return a
}

func maxUint(a, b uint) uint {
	if a < b {
		return b
	}
	return a
}
func maxInt64(a, b int64) int64 {
	if a < b {
		return b
	}
	return a
}

func maxUint64(a, b uint64) uint64 {
	if a < b {
		return b
	}
	return a
}

func absInt(a int) int {
	if a < 0 {
		return -a
	}
	return a
}

func absInt64(a int64) int64 {
	if a < 0 {
		return -a
	}
	return a
}

Score : 400 points

Problem Statement

We have a sequence of length N, a = (a_1, a_2, ..., a_N).
Each a_i is a positive integer.

Snuke's objective is to permute the element in a so that the following condition is satisfied:

For each 1 ≤ i ≤ N - 1, the product of a_i and a_{i + 1} is a multiple of 4.

Determine whether Snuke can achieve his objective.

Constraints

2 ≤ N ≤ 10^5

a_i is an integer.

1 ≤ a_i ≤ 10^9

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

If Snuke can achieve his objective, print Yes; otherwise, print No.

Sample Input 1

3
1 10 100

Sample Output 1

Yes

One solution is (1, 100, 10).

Sample Input 2

4
1 2 3 4

Sample Output 2

No

It is impossible to permute a so that the condition is satisfied.

Sample Input 3

3
1 4 1

Sample Output 3

Yes

The condition is already satisfied initially.

Sample Input 4

2
1 1

Sample Output 4

No

Sample Input 5

6
2 7 1 8 2 8

Sample Output 5

Yes


package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
	"strings"
)

var sc = bufio.NewScanner(os.Stdin)

func stringLineScan() string {
	sc.Scan()
	return sc.Text()
}

func main() {
	sc.Split(bufio.ScanWords)
	h, _ := strconv.Atoi(stringLineScan())
	w, _ := strconv.Atoi(stringLineScan())
	n, _ := strconv.Atoi(stringLineScan())
	c := 0
	ans := make([][]int, n)
	for i := 0; i < h; i++ {
		t := make([]int, w)
		ans[i] = t
	}
	for i := 1; i <= n; i++ {
		t, _ := strconv.Atoi(stringLineScan())
		for j := 0; j < t; j++ {
			ans[c/w][c%w] = i
			c++
		}
	}
	for i := 0; i < h; i++ {
		if i%2 == 0 {
			fmt.Println(strings.Trim(fmt.Sprint(ans[i]), "[]"))
		} else {
			t := make([]int, w)
			for j := 0; j < w; j++ {
				t[w-j-1] = ans[i][j]
			}
			fmt.Println(strings.Trim(fmt.Sprint(t), "[]"))
		}
	}
}

Score : 400 points

Problem Statement

We have a grid with H rows and W columns of squares.
Snuke is painting these squares in colors 1, 2, ..., N.
Here, the following conditions should be satisfied:

For each i (1 ≤ i ≤ N), there are exactly a_i squares painted in Color i. Here, a_1 + a_2 + ... + a_N = H W.

For each i (1 ≤ i ≤ N), the squares painted in Color i are 4-connected. That is, every square painted in Color i can be reached from every square painted in Color i by repeatedly traveling to a horizontally or vertically adjacent square painted in Color i.

Find a way to paint the squares so that the conditions are satisfied.
It can be shown that a solution always exists.

Constraints

1 ≤ H, W ≤ 100

1 ≤ N ≤ H W

a_i ≥ 1

a_1 + a_2 + ... + a_N = H W

Input

Input is given from Standard Input in the following format:

H W
N
a_1 a_2 ... a_N

Output

Print one way to paint the squares that satisfies the conditions.
Output in the following format:

c_{1 1} ... c_{1 W}
:
c_{H 1} ... c_{H W}

Here, c_{i j} is the color of the square at the i-th row from the top and j-th column from the left.

Sample Input 1

2 2
3
2 1 1

Sample Output 1

1 1
2 3

Below is an example of an invalid solution:

1 2
3 1

This is because the squares painted in Color 1 are not 4-connected.

Sample Input 2

3 5
5
1 2 3 4 5

Sample Output 2

1 4 4 4 3
2 5 4 5 3
2 5 5 5 3

Sample Input 3

1 1
1
1

Sample Output 3

1


// Package main provides
//
// File:  c.go
// Author: ymiyamoto
//
// Created on Thu Jan 10 09:47:06 2019
//
package main

import "fmt"

func main() {
	var H, W, N int
	fmt.Scan(&H, &W, &N)
	arr := make([]int, N)
	for i := range arr {
		fmt.Scan(&arr[i])
	}

	board := make([][]int, H)
	for i := range board {
		board[i] = make([]int, W)
	}

	h, w := 0, 0
	for i := range arr {
		for arr[i] > 0 {
			board[h][w] = i + 1
			w++
			if w == W {
				h++
				w = 0
			}
			arr[i]--
		}
	}

	for i := range board {
		if i%2 == 0 {
			for j := range board[i] {
				if j != 0 {
					fmt.Print(" ")
				}
				fmt.Print(board[i][j])
			}
		} else {
			for j := len(board[i]) - 1; j >= 0; j-- {
				if j != len(board[i])-1 {
					fmt.Print(" ")
				}
				fmt.Print(board[i][j])
			}
		}
		fmt.Println()
	}
}


package main

import (
	"bufio"
	"fmt"
	"os"
)

func main() {
	K := ReadInt()
	N := 50
	r := K % N
	A := (K + (N - 1)) / N
	B := K / N
	v := make([]int, N)
	for i := 0; i < r; i++ {
		v[i] = (N - 1) - (K - B - 1) + A*N
	}
	for i := r; i < N; i++ {
		v[i] = (N - 1) - (K - B) + B*N
	}
	fmt.Println(len(v))
	for i := 0; i < N; i++ {
		fmt.Printf("%d ", v[i])
	}
	fmt.Println()
}

var reader = bufio.NewReader(os.Stdin)

func Scan(a ...interface{}) {
	if _, err := fmt.Fscan(reader, a...); err != nil {
		panic(err)
	}
}
func ReadInt() (i int)       { Scan(&i); return }
func ReadString() (s string) { Scan(&s); return }
func ReadInts(n int) []int {
	v := make([]int, n)
	for i := 0; i < n; i++ {
		Scan(&v[i])
	}
	return v
}

Score : 600 points

Problem Statement

We have a sequence of length N consisting of non-negative integers. Consider performing the following operation on this sequence until the largest element in this sequence becomes N-1 or smaller.

Determine the largest element in the sequence (if there is more than one, choose one). Decrease the value of this element by N, and increase each of the other elements by 1.

It can be proved that the largest element in the sequence becomes N-1 or smaller after a finite number of operations.

You are given an integer K. Find an integer sequence a_i such that the number of times we will perform the above operation is exactly K. It can be shown that there is always such a sequence under the constraints on input and output in this problem.

Constraints

0 ≤ K ≤ 50 \times 10^{16}

Input

Input is given from Standard Input in the following format:

K

Output

Print a solution in the following format:

N
a_1 a_2 ... a_N

Here, 2 ≤ N ≤ 50 and 0 ≤ a_i ≤ 10^{16} + 1000 must hold.

Sample Input 1

0

Sample Output 1

4
3 3 3 3

Sample Input 2

1

Sample Output 2

3
1 0 3

Sample Input 3

2

Sample Output 3

2
2 2

The operation will be performed twice: [2, 2] -> [0, 3] -> [1, 1].

Sample Input 4

3

Sample Output 4

7
27 0 0 0 0 0 0

Sample Input 5

1234567894848

Sample Output 5

10
1000 193 256 777 0 1 1192 1234567891011 48 425


package main

import (
	"bufio"
	"fmt"
	"math"
	"os"
	"reflect"
	"sort"
	"strconv"
	"strings"
	// "regexp"
)

/* sort */
type pair struct {
	index  int64
	p1, p2 interface{}
}
type pairs []pair

func (slice pairs) Len() int {
	return len(slice)
}

func (slice pairs) Less(i, j int) bool {
	return slice[i].index < slice[j].index
}

func (slice pairs) Swap(i, j int) {
	slice[i], slice[j] = slice[j], slice[i]
}

/* sort */
type Int64Slice []int64

func (slice Int64Slice) Len() int {
	return len(slice)
}

func (slice Int64Slice) Less(i, j int) bool {
	return slice[i] < slice[j]
}

func (slice Int64Slice) Swap(i, j int) {
	slice[i], slice[j] = slice[j], slice[i]
}
func Int64s(a []int64)                    { sort.Sort(Int64Slice(a)) }
func Int64sSliceAreSorted(a []int64) bool { return sort.IsSorted(Int64Slice(a)) }

const (
	initialBufSize = 1e4
	maxBufSize     = 1e8
	INF            = 1e8
)

var (
	scanner = bufio.NewScanner(os.Stdin)
	writer  = bufio.NewWriter(os.Stdout)
	di      = []int64{-1, 0, 1, 0}
	dj      = []int64{0, -1, 0, 1}
	di8     = []int64{-1, -1, 0, 1, 1, 1, 0, -1}
	dj8     = []int64{0, -1, -1, -1, 0, 1, 1, 1}
)

type edge struct {
	s, t int64
}

func main() {
	buf := make([]byte, initialBufSize)
	scanner.Buffer(buf, maxBufSize)
	scanner.Split(bufio.ScanWords)
	N, M := readInt(), readInt()
	edges := make(map[edge]int64, N+1)
	for i := int64(0); i < M; i++ {
		a, b := readInt()-1, readInt()-1
		edges[edge{a, b}]++
		edges[edge{b, a}]++
	}
	for i := int64(0); i < M; i++ {
		if edges[edge{0, i}] > 0 && edges[edge{i, N - 1}] > 0 {
			fmt.Println("POSSIBLE")
			return
		}
	}
	fmt.Println("IMPOSSIBLE")
}

/*==========================================
 *             Library
 *==========================================*/
func NextPermutation(x sort.Interface) bool {
	n := x.Len() - 1
	if n < 1 {
		return false
	}
	j := n - 1
	for ; !x.Less(j, j+1); j-- {
		if j == 0 {
			return false
		}
	}
	l := n
	for !x.Less(j, l) {
		l--
	}
	x.Swap(j, l)
	for k, l := j+1, n; k < l; {
		x.Swap(k, l)
		k++
		l--
	}
	return true
}
func doubleAry(h int64, w int64, init int64) (res [][]int64) {
	res = make([][]int64, h)
	for i := int64(0); i < h; i++ {
		res[i] = make([]int64, w)
		for j := int64(0); j < w; j++ {
			res[i][j] = init
		}
	}
	return
}

func aryEq(a []int64, b []int64) bool {
	return reflect.DeepEqual(a, b)
}

func clone(n []int64) (r []int64) {
	r = make([]int64, len(n))
	for i := 0; i < len(n); i++ {
		r[i] = n[i]
	}
	return
}
func write(s string) {
	writer.WriteString(s)
}

func print() {
	writer.Flush()
}

// scanner.Split(bufio.ScanWords) をコメントアウトしないと使用不可
func readLine() (s string) {
	if scanner.Scan() {
		s = scanner.Text()
	}
	return s
}

func readInt() (read int64) {
	scanner.Scan()
	read, err := strconv.ParseInt(scanner.Text(), 10, 64)
	if err != nil {
		panic(err)
	}
	return
}

func readFloat() (read float64) {
	scanner.Scan()
	read, err := strconv.ParseFloat(scanner.Text(), 64)
	if err != nil {
		panic(err)
	}
	return
}

func readRunes() (read []rune) {
	scanner.Scan()
	for _, v := range scanner.Text() {
		read = append(read, v)
	}
	return
}

func readString() (read string) {
	scanner.Scan()
	read = scanner.Text()
	return
}

func readStrings() (read []string) {
	scanner.Scan()
	for _, v := range scanner.Text() {
		read = append(read, string(v))
	}
	return
}

func s2i(s string) int64 {
	var intVal, e = strconv.ParseInt(s, 10, 64)
	if e != nil {
		panic(e)
	}
	return intVal
}

func i2s(i int64) string {
	var strVal = strconv.FormatInt(i, 10)
	return strVal
}

func s2f(s string) float64 {
	var floatVal, e = strconv.ParseFloat(s, 64)
	if e != nil {
		panic(e)
	}
	return floatVal
}

func abs(i int64) int64 {
	return int64(math.Abs(float64(i)))
}

func max(a ...int64) int64 {
	ans := int64(0)
	for i, v := range a {
		if i == 0 {
			ans = v
		}
		if ans < v {
			ans = v
		}
	}
	return ans
}
func min(a ...int64) int64 {
	ans := int64(0)
	for i, v := range a {
		if i == 0 {
			ans = v
		}
		if ans > v {
			ans = v
		}
	}
	return ans
}

func sum(i []int64) int64 {
	sum := int64(0)
	for _, val := range i {
		sum += val
	}
	return sum
}

func split(s string) []string {
	return strings.Fields(s)
}

func strAry2intAry(strs []string) []int64 {
	var ret = make([]int64, 0, len(strs))
	for _, str := range strs {
		var intVal = s2i(str)
		ret = append(ret, intVal)
	}
	return ret
}

func intAry2strAry(nums []int64) []string {
	var ret = make([]string, 0, len(nums))
	for _, num := range nums {
		var strVal = i2s(num)
		ret = append(ret, strVal)
	}
	return ret
}

func ary2str(strs []string) string {
	return strings.Join(strs, " ")
}

func reverse(res []int64) []int64 {
	for i, j := 0, len(res)-1; i < j; i, j = i+1, j-1 {
		res[i], res[j] = res[j], res[i]
	}
	return res
}
func reverseStr(res []string) []string {
	for i, j := 0, len(res)-1; i < j; i, j = i+1, j-1 {
		res[i], res[j] = res[j], res[i]
	}
	return res
}

func ini(res []int, init int) {
	if len(res) == 0 {
		return
	}
	res[0] = init
	for i := 0; i < len(res); i++ {
		copy(res[i:], res[:i])
	}
}

//
// func regexpExample() {
//     // Your code here!
//     var str = "13:20"
//     r := regexp.MustCompile(`(\d+):(\d+)`)
//     fmt.Println(r.FindStringSubmatch(str))
// }

// type Country struct {
//     gold int
//     silver int
//     blonze int
// }

// // 複数ソートのサンプル
// func stableSortExample() []Country{
//     var country = []Country {
//         {gold:1, silver:2, blonze:3},
//         {gold:1, silver:2, blonze:3},
//         {gold:1, silver:3, blonze:2},
//         {gold:1, silver:3, blonze:3},
//     }
//     sort.SliceStable(country, func(i, j int) bool { return country[i].blonze > country[j].blonze })
//     sort.SliceStable(country, func(i, j int) bool { return country[i].silver > country[j].silver })
//     sort.SliceStable(country, func(i, j int) bool { return country[i].gold > country[j].gold })
//     return country
// }

Score : 300 points

Problem Statement

In Takahashi Kingdom, there is an archipelago of N islands, called Takahashi Islands.
For convenience, we will call them Island 1, Island 2, ..., Island N.

There are M kinds of regular boat services between these islands.
Each service connects two islands. The i-th service connects Island a_i and Island b_i.

Cat Snuke is on Island 1 now, and wants to go to Island N.
However, it turned out that there is no boat service from Island 1 to Island N, so he wants to know whether it is possible to go to Island N by using two boat services.

Help him.

Constraints

3 ≤ N ≤ 200 000

1 ≤ M ≤ 200 000

1 ≤ a_i < b_i ≤ N

(a_i, b_i) \neq (1, N)

If i \neq j, (a_i, b_i) \neq (a_j, b_j).

Input

Input is given from Standard Input in the following format:

N M
a_1 b_1
a_2 b_2
:
a_M b_M

Output

If it is possible to go to Island N by using two boat services, print POSSIBLE; otherwise, print IMPOSSIBLE.

Sample Input 1

3 2
1 2
2 3

Sample Output 1

POSSIBLE

Sample Input 2

4 3
1 2
2 3
3 4

Sample Output 2

IMPOSSIBLE

You have to use three boat services to get to Island 4.

Sample Input 3

100000 1
1 99999

Sample Output 3

IMPOSSIBLE

Sample Input 4

5 5
1 3
4 5
2 3
2 4
1 4

Sample Output 4

POSSIBLE

You can get to Island 5 by using two boat services: Island 1 -> Island 4 -> Island 5.


package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
)

func main() {
	scn := newScanner()
	N := scn.nextInt()
	K := scn.nextInt()
	A := make([]int, N)
	maxA := 0
	for i := 0; i < N; i++ {
		A[i] = scn.nextInt()
		maxA = max(maxA, A[i])
	}
	flag := false
	for i := 0; i < N-1; i++ {
		g := gcd(A[i], A[i+1])
		if K%g == 0 {
			flag = true
		}
	}
	if flag && maxA > K {
		fmt.Println("POSSIBLE")
	} else {
		fmt.Println("IMPOSSIBLE")
	}
}

func max(a ...int) int {
	r := a[0]
	for i := 0; i < len(a); i++ {
		if r < a[i] {
			r = a[i]
		}
	}
	return r
}

func gcd(a, b int) int {
	for b != 0 {
		a, b = b, a%b
	}
	return a
}

type scanner struct {
	r   *bufio.Reader
	buf []byte
	p   int
}

func newScanner() *scanner {
	rdr := bufio.NewReaderSize(os.Stdin, 10000)
	return &scanner{r: rdr}
}
func (s *scanner) next() string {
	s.pre()
	start := s.p
	for ; s.p < len(s.buf); s.p++ {
		if s.buf[s.p] == ' ' {
			break
		}
	}
	result := string(s.buf[start:s.p])
	s.p++
	return result
}
func (s *scanner) nextLine() string {
	s.pre()
	start := s.p
	s.p = len(s.buf)
	return string(s.buf[start:])
}
func (s *scanner) nextInt() int {
	v, _ := strconv.Atoi(s.next())
	return v
}
func (s *scanner) nextInt64() int64 {
	v, _ := strconv.ParseInt(s.next(), 10, 64)
	return v
}
func (s *scanner) pre() {
	if s.p >= len(s.buf) {
		s.readLine()
		s.p = 0
	}
}
func (s *scanner) readLine() {
	s.buf = make([]byte, 0)
	for {
		l, p, e := s.r.ReadLine()
		if e != nil {
			panic(e)
		}
		s.buf = append(s.buf, l...)
		if !p {
			break
		}
	}
}

Score : 300 points

Problem Statement

There is a box containing N balls. The i-th ball has the integer A_i written on it.
Snuke can perform the following operation any number of times:

Take out two balls from the box. Then, return them to the box along with a new ball, on which the absolute difference of the integers written on the two balls is written.

Determine whether it is possible for Snuke to reach the state where the box contains a ball on which the integer K is written.

Constraints

1 \leq N \leq 10^5

1 \leq A_i \leq 10^9

1 \leq K \leq 10^9

All input values are integers.

Input

Input is given from Standard Input in the following format:

N K
A_1 A_2 ... A_N

Output

If it is possible for Snuke to reach the state where the box contains a ball on which the integer K is written, print POSSIBLE; if it is not possible, print IMPOSSIBLE.

Sample Input 1

3 7
9 3 4

Sample Output 1

POSSIBLE

First, take out the two balls 9 and 4, and return them back along with a new ball, abs(9-4)=5.
Next, take out 3 and 5, and return them back along with abs(3-5)=2.
Finally, take out 9 and 2, and return them back along with abs(9-2)=7.
Now we have 7 in the box, and the answer is therefore POSSIBLE.

Sample Input 2

3 5
6 9 3

Sample Output 2

IMPOSSIBLE

No matter what we do, it is not possible to have 5 in the box. The answer is therefore IMPOSSIBLE.

Sample Input 3

4 11
11 3 7 15

Sample Output 3

POSSIBLE

The box already contains 11 before we do anything. The answer is therefore POSSIBLE.

Sample Input 4

5 12
10 2 8 6 4

Sample Output 4

IMPOSSIBLE


package main

import (
        "fmt"
        "sort"
)

func main () {
        var n, k, l, an int
        f := []int{}
        fmt.Scan(&n,&k)
        for i := 0 ; i < n ; i++ {
                fmt.Scan(&l)
                f = append (f, l)
        }
        sort.Sort(sort.IntSlice(f))
        sort.Sort(sort.Reverse(sort.IntSlice(f)))
        for i := 0 ; i < k ; i++{
                an += f[i]
                }
        fmt.Println(an)
}

Score : 200 points

Problem Statement

Snuke has N sticks.
The length of the i-th stick is l_i.

Snuke is making a snake toy by joining K of the sticks together.

The length of the toy is represented by the sum of the individual sticks that compose it.
Find the maximum possible length of the toy.

Constraints

1 \leq K \leq N \leq 50

1 \leq l_i \leq 50

l_i is an integer.

Input

Input is given from Standard Input in the following format:

N K
l_1 l_2 l_3 ... l_{N}

Output

Print the answer.

Sample Input 1

5 3
1 2 3 4 5

Sample Output 1

12

You can make a toy of length 12 by joining the sticks of lengths 3, 4 and 5, which is the maximum possible length.

Sample Input 2

15 14
50 26 27 21 41 7 42 35 7 5 5 36 39 1 45

Sample Output 2

386


package main

import (
	"bufio"
	"fmt"
	"math"
	"os"
	"strconv"
)

//Util
func nextStr() string {
	sc.Scan()
	return sc.Text()
}

func nextInt() int {
	sc.Scan()
	i, e := strconv.Atoi(sc.Text())
	if e != nil {
		panic(e)
	}
	return i
}

func maxInt(a, b int) int {
	if a > b {
		return a
	} else {
		return b
	}
}

func minInt(a, b int) int {
	if a > b {
		return b
	} else {
		return a
	}
}

func absInt(a int) int {
	if a >= 0 {
		return a
	} else {
		return -a
	}
}

func absFloat64(a float64) float64 {
	if a >= 0 {
		return a
	} else {
		return -a
	}
}

func maxFloat64(a, b float64) float64 {
	if a > b {
		return a
	} else {
		return b
	}
}

func minFloat64(a, b float64) float64 {
	if a > b {
		return b
	} else {
		return a
	}
}

func str2Int(s string) int {
	n, _ := strconv.Atoi(s)
	return n
}

func reverse(s string) string {
	runes := []rune(s)
	for i, j := 0, len(runes)-1; i < j; i, j = i+1, j-1 {
		runes[i], runes[j] = runes[j], runes[i]
	}
	return string(runes)
}

func powInt(p, q int) int {
	return int(math.Pow(float64(p), float64(q)))
}

func isPrime(x int) bool {
	if x == 1 {
		return false
	}
	if x == 2 {
		return true
	}
	if x%2 == 0 {
		return false
	}

	b := true
	rootx := int(math.Sqrt(float64(x)))
	i := 3
	for i <= rootx {
		if x%i == 0 {
			b = false
			break
		}
		i += 2
	}
	return b
}

func PrimeFactors(n int) (pfs []int) {
	// Get the number of 2s that divide n
	for n%2 == 0 {
		pfs = append(pfs, 2)
		n = n / 2
	}

	// n must be odd at this point. so we can skip one element
	// (note i = i + 2)
	for i := 3; i*i <= n; i = i + 2 {
		// while i divides n, append i and divide n
		for n%i == 0 {
			pfs = append(pfs, i)
			n = n / i
		}
	}

	// This condition is to handle the case when n is a prime number
	// greater than 2
	if n > 2 {
		pfs = append(pfs, n)
	}

	return
}

func PrimeFactorsMap(n int) map[int]int {
	pfs := map[int]int{}
	// Get the number of 2s that divide n
	for n%2 == 0 {
		if _, ok := pfs[2]; ok {
			pfs[2]++
		} else {
			pfs[2] = 1
		}
		//pfs = append(pfs, 2)
		n = n / 2
	}

	// n must be odd at this point. so we can skip one element
	// (note i = i + 2)
	for i := 3; i*i <= n; i = i + 2 {
		// while i divides n, append i and divide n
		for n%i == 0 {
			if _, ok := pfs[i]; ok {
				pfs[i]++
			} else {
				pfs[i] = 1
			}
			n = n / i
		}
	}

	// This condition is to handle the case when n is a prime number
	// greater than 2
	if n > 2 {
		if _, ok := pfs[n]; ok {
			pfs[n]++
		} else {
			pfs[n] = 1
		}
	}

	return pfs
}

func sumInts(x []int) int {

	total := 0
	for _, v := range x {
		total += v
	}

	return total
}

func gcd(a, b int) int {
	if b == 0 {
		return a
	}
	return gcd(b, a%b)
}

func lcm(x, y int) int {

	return x * y / gcd(x, y)

}

//Main
const (
	initialBufSize = 10000
	maxBufSize     = 1000000
	mod            = 1e9 + 7
)

var (
	sc *bufio.Scanner = func() *bufio.Scanner {
		sc := bufio.NewScanner(os.Stdin)
		buf := make([]byte, initialBufSize)
		sc.Buffer(buf, maxBufSize)
		return sc
	}()
)

func main() {
	sc.Split(bufio.ScanWords)

	N, P := nextInt(), nextInt()
	odd := 0
	even := 0
	for i := 0; i < N; i++ {
		A := nextInt()
		if A%2 == 0 {
			even++
		} else {
			odd++
		}
	}

	cnb := make([]int, maxInt(odd+1, 2))
	cnb[0] = 1
	cnb[1] = odd
	for i := 2; i <= odd; i++ {
		cnb[i] = (cnb[i-1] * (odd - i + 1)) / i
	}

	if P == 0 {
		ans := 0
		e := powInt(2, even)
		for i := 0; i <= odd; i += 2 {
			ans += e * cnb[i]
		}
		fmt.Println(ans)
	} else {
		ans := 0
		e := powInt(2, even)
		for i := 1; i < odd; i += 2 {
			ans += e * cnb[i]
		}
		fmt.Println(ans)
	}

	//fmt.Println(cnb, len(cnb), odd, even)

}

Score : 200 points

Problem Statement

There are N bags of biscuits. The i-th bag contains A_i biscuits.

Takaki will select some of these bags and eat all of the biscuits inside.
Here, it is also possible to select all or none of the bags.

He would like to select bags so that the total number of biscuits inside is congruent to P modulo 2.
How many such ways to select bags there are?

Constraints

1 \leq N \leq 50

P = 0 or 1

1 \leq A_i \leq 100

Input

Input is given from Standard Input in the following format:

N P
A_1 A_2 ... A_N

Output

Print the number of ways to select bags so that the total number of biscuits inside is congruent to P modulo 2.

Sample Input 1

2 0
1 3

Sample Output 1

2

There are two ways to select bags so that the total number of biscuits inside is congruent to 0 modulo 2:

Select neither bag. The total number of biscuits is 0.

Select both bags. The total number of biscuits is 4.

Sample Input 2

1 1
50

Sample Output 2

0

Sample Input 3

3 0
1 1 1

Sample Output 3

4

Two bags are distinguished even if they contain the same number of biscuits.

Sample Input 4

45 1
17 55 85 55 74 20 90 67 40 70 39 89 91 50 16 24 14 43 24 66 25 9 89 71 41 16 53 13 61 15 85 72 62 67 42 26 36 66 4 87 59 91 4 25 26

Sample Output 4

17592186044416


package main

import "fmt"

func main() {
	var n, p, a, odd, even int
	fmt.Scan(&n, &p)

	for i := 0; i < n; i++ {
		fmt.Scan(&a)
		odd += a % 2
		even += (a % 2) ^ 1
	}
	if odd == 0 {
		fmt.Println(pow(2, even) * toInt(p == 0))
	} else {
		fmt.Println(pow(2, odd-1) * pow(2, even))
	}
}

func pow(a, b int) int {
	result := 1
	for i := 0; i < b; i++ {
		result *= a
	}
	return result
}

func toInt(ok bool) int {
	if ok {
		return 1
	}
	return 0
}


package main

import(
	"bufio"
	"os"
	"strconv"
	"fmt"
)

var sc = bufio.NewScanner(os.Stdin)

func nextInt() int {
	sc.Scan()
	i, e := strconv.Atoi(sc.Text())
	if e != nil {
		panic(e)
	}
	return i
}

func main() {
	sc.Split(bufio.ScanWords)
	x := []int{}
	var max = 0
	var sum = 0
	for i := 0; i < 3; i++ {
		x = append(x, nextInt())
		sum += x[i]
		if (x[i] > max) {
			max = x[i]
		}
	}
	fmt.Println(sum - max)
}

Score : 100 points

Problem Statement

Snuke is buying a bicycle.
The bicycle of his choice does not come with a bell, so he has to buy one separately.

He has very high awareness of safety, and decides to buy two bells, one for each hand.

The store sells three kinds of bells for the price of a, b and c yen (the currency of Japan), respectively.
Find the minimum total price of two different bells.

Constraints

1 \leq a,b,c \leq 10000

a, b and c are integers.

Input

Input is given from Standard Input in the following format:

a b c

Output

Print the minimum total price of two different bells.

Sample Input 1

700 600 780

Sample Output 1

1300

Buying a 700-yen bell and a 600-yen bell costs 1300 yen.

Buying a 700-yen bell and a 780-yen bell costs 1480 yen.

Buying a 600-yen bell and a 780-yen bell costs 1380 yen.

The minimum among these is 1300 yen.

Sample Input 2

10000 10000 10000

Sample Output 2

20000

Buying any two bells costs 20000 yen.


package main

import (
	"bufio"
	"fmt"
	"log"
	"os"
	"strconv"
)

// const abcd = "abcdefghijklmnopqrstuvwxyz"
// const ABCD = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"

// var dy = [...]int{1, 1, 0, -1, -1, -1, 0, 1}
// var dx = [...]int{0, 1, 1, 1, 0, -1, -1, -1}

// var dx = [...]int{0, 1, 0, -1}
// var dy = [...]int{1, 0, -1, 0}

var inf int = 1e13
var mod = 1000000007

// ---------------------------------------------------------
func main() {
	log.SetFlags(log.Lshortfile)
	next := newScanner()
	s := next.next()
	ans := 0
	for i := len(s); i >= 0; i-- {
		if check(s[:i]) {
			ans = len(s[:i]) / 2
			break
		}
	}

	fmt.Println(ans)
}

func check(s string) bool {
	if len(s)%2 != 0 {
		return false
	}
	for i := 0; i >= len(s)/2; i++ {
		if s[i] != s[len(s)-i] {
			return false
		}
	}
	return true
}

// ---------------------------------------------------------

// Pair is liked C++ pair
type Pair struct {
	a, b int
}

// Pairs is sorted by []Pair struct
type Pairs []Pair

func (p Pairs) Len() int {
	return len(p)
}
func (p Pairs) Swap(i, j int) {
	p[i], p[j] = p[j], p[i]
}

func (p Pairs) Less(i, j int) bool {
	if p[i].a < p[j].a {
		return true
	} else if p[i].a == p[j].a {
		return p[i].b < p[j].b
	}
	return false
}

// -------------------------------
func in(c, a, z int) bool {
	return c >= a && c < z
}

func btoi(b bool) int {
	if b {
		return 1
	}
	return 0
}

func itob(a int) bool {
	if a == 0 {
		return false
	}
	return true
}

func max(a ...int) int {
	r := a[0]
	for i := 0; i < len(a); i++ {
		if r < a[i] {
			r = a[i]
		}
	}
	return r
}

func sum(a ...int) int {
	r := a[0]
	if len(a) > 1 {
		for i := 1; i < len(a); i++ {
			r += a[i]
		}
	}
	return r
}

func min(a ...int) int {
	r := a[0]
	for i := 0; i < len(a); i++ {
		if r > a[i] {
			r = a[i]
		}
	}
	return r
}

func minmax(a, b int) (int, int) {
	if a > b {
		return b, a
	}
	return a, b
}

func abs(a int) int {
	if a < 0 {
		return -a
	}
	return a
}

// ---------- buffered scanner -----------------------------------------
type scanner struct {
	r   *bufio.Reader
	buf []byte
	p   int
}

func newScanner() *scanner {
	rdr := bufio.NewReaderSize(os.Stdin, 10000)
	return &scanner{r: rdr}
}
func (s *scanner) next() string {
	s.pre()
	start := s.p
	for ; s.p < len(s.buf); s.p++ {
		if s.buf[s.p] == ' ' {
			break
		}
	}
	result := string(s.buf[start:s.p])
	s.p++
	return result
}
func (s *scanner) Line() string {
	s.pre()
	start := s.p
	s.p = len(s.buf)
	return string(s.buf[start:])
}
func (s *scanner) Int() int {
	v, _ := strconv.Atoi(s.next())
	return v
}
func (s *scanner) Int64() int64 {
	v, _ := strconv.ParseInt(s.next(), 10, 64)
	return v
}
func (s *scanner) pre() {
	if s.p >= len(s.buf) {
		s.readLine()
		s.p = 0
	}
}
func (s *scanner) readLine() {
	s.buf = make([]byte, 0)
	for {
		l, p, e := s.r.ReadLine()
		if e != nil {
			panic(e)
		}
		s.buf = append(s.buf, l...)
		if !p {
			break
		}
	}
}

Score : 200 points

Problem Statement

We will call a string that can be obtained by concatenating two equal strings an even string.
For example, xyzxyz and aaaaaa are even, while ababab and xyzxy are not.

You are given an even string S consisting of lowercase English letters.
Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S.
It is guaranteed that such a non-empty string exists for a given input.

Constraints

2 \leq |S| \leq 200

S is an even string consisting of lowercase English letters.

There exists a non-empty even string that can be obtained by deleting one or more characters from the end of S.

Input

Input is given from Standard Input in the following format:

S

Output

Print the length of the longest even string that can be obtained.

Sample Input 1

abaababaab

Sample Output 1

6

abaababaab itself is even, but we need to delete at least one character.

abaababaa is not even.

abaababa is not even.

abaabab is not even.

abaaba is even. Thus, we should print its length, 6.

Sample Input 2

xxxx

Sample Output 2

2

xxx is not even.

xx is even.

Sample Input 3

abcabcabcabc

Sample Output 3

6

The longest even string that can be obtained is abcabc, whose length is 6.

Sample Input 4

akasakaakasakasakaakas

Sample Output 4

14

The longest even string that can be obtained is akasakaakasaka, whose length is 14.


package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
	"strings"
)

func main() {
	sc := bufio.NewScanner(os.Stdin)
	sc.Scan()
	ns := sc.Text()
	sc.Scan()
	line := strings.Split(sc.Text(), " ")
	n, _ := strconv.Atoi(ns)
	ad := n % 2
	l, r := make([]string, n / 2), make([]string, (n / 2) + ad)
	for i, v := range line {
		if i % 2 == 0 {
			r[i / 2] = v
		} else {
			l[i / 2] = v
		}
	}
	if n % 2 == 0 {
		for i := len(l) - 1; i >= 0; i-- {
			fmt.Print(l[i])
			if i != 0 {
				fmt.Print(" ")
			}
		}
		for i := 0; i < len(r); i++ {
			fmt.Print(" ")
			fmt.Print(r[i])
		}
	} else {
		for i := len(r) - 1; i >= 0; i-- {
			fmt.Print(r[i])
			if i != 0 {
				fmt.Print(" ")
			}
		}
		for i := 0; i < len(l); i++ {
			fmt.Print(" ")
			fmt.Print(l[i])
		}
	}
	fmt.Println()
}

Score : 300 points

Problem Statement

You are given an integer sequence of length n, a_1, ..., a_n.
Let us consider performing the following n operations on an empty sequence b.

The i-th operation is as follows:

Append a_i to the end of b.

Reverse the order of the elements in b.

Find the sequence b obtained after these n operations.

Constraints

1 \leq n \leq 2\times 10^5

0 \leq a_i \leq 10^9

n and a_i are integers.

Input

Input is given from Standard Input in the following format:

n
a_1 a_2 ... a_n

Output

Print n integers in a line with spaces in between.
The i-th integer should be b_i.

Sample Input 1

4
1 2 3 4

Sample Output 1

4 2 1 3

After step 1 of the first operation, b becomes: 1.

After step 2 of the first operation, b becomes: 1.

After step 1 of the second operation, b becomes: 1, 2.

After step 2 of the second operation, b becomes: 2, 1.

After step 1 of the third operation, b becomes: 2, 1, 3.

After step 2 of the third operation, b becomes: 3, 1, 2.

After step 1 of the fourth operation, b becomes: 3, 1, 2, 4.

After step 2 of the fourth operation, b becomes: 4, 2, 1, 3.

Thus, the answer is 4 2 1 3.

Sample Input 2

3
1 2 3

Sample Output 2

3 1 2

As shown above in Sample Output 1, b becomes 3, 1, 2 after step 2 of the third operation. Thus, the answer is 3 1 2.

Sample Input 3

1
1000000000

Sample Output 3

1000000000

Sample Input 4

6
0 6 7 6 7 0

Sample Output 4

0 6 6 0 7 7


package main

import (
    "bufio"
    "os"
    "strconv"
    "fmt"
)

const MOD = 1000000007

/* input */
var sc = bufio.NewScanner(os.Stdin)

func readInt() int {
    sc.Scan()
    n, _ := strconv.Atoi(sc.Text())
    return n
}

/* fact */
func fact(x int, factMem map[int]int) int {
    if x == 0 {
        return 1
    }
    value, ok := factMem[x]
    if ok {
        return value
    }
    factMem[x] = x * fact(x - 1, factMem) % MOD
    return factMem[x]
}

/* inverse */
func inverse(x int, factMem map[int]int, inverseMem map[int]int) int {
    value, ok := inverseMem[x]
    if ok {
        return value
    }
    inverseMem[x] = power(fact(x, factMem), MOD - 2, MOD)
    return inverseMem[x]
}

/* power */
func power(x, y, z int) int {
    if y == 0 {
        return 1
    }
    if y % 2 == 0 {
        return power(x * x % z, y / 2, z)
    } else {
        return x * power(x, y - 1, z) % z
    }
    return 0
}

/* comb */
func comb(x int, y int, factMem map[int]int, inverseMem map[int]int) int {
    if x < y || y < 0 {
        return 0
    }
    return fact(x, factMem) * inverse(y, factMem, inverseMem) % MOD * inverse(x - y, factMem, inverseMem) % MOD
}

func main() {
    sc.Split(bufio.ScanWords)
    n := readInt()
    dic := make(map[int]int)
    left, right := 0, 0
    for i := 0; i < n + 1; i++ {
        a := readInt()
        index, ok := dic[a]
        if ok {
            left = index
            right = i
            break
        } else {
            dic[a] = i
        }

    }

    rem := n - right + left
    factMem := make(map[int]int)
    inverseMem := make(map[int]int)
    for i := 0; i < n + 1; i++ {
        fact(i, factMem)
    }
    wrt := bufio.NewWriter(os.Stdout)
    for i := 0; i < n + 1; i++ {
        fmt.Fprintln(wrt, (MOD + comb(n + 1, i + 1, factMem, inverseMem) - comb(rem, i, factMem, inverseMem)) % MOD)
        wrt.Flush()
    }
}

Score : 600 points

Problem Statement

You are given an integer sequence of length n+1, a_1,a_2,...,a_{n+1}, which consists of the n integers 1,...,n.
It is known that each of the n integers 1,...,n appears at least once in this sequence.

For each integer k=1,...,n+1, find the number of the different subsequences (not necessarily contiguous) of the given sequence with length k, modulo 10^9+7.

Notes

If the contents of two subsequences are the same, they are not separately counted even if they originate from different positions in the original sequence.

A subsequence of a sequence a with length k is a sequence obtained by selecting k of the elements of a and arranging them without changing their relative order. For example, the sequences 1,3,5 and 1,2,3 are subsequences of 1,2,3,4,5, while 3,1,2 and 1,10,100 are not.

Constraints

1 \leq n \leq 10^5

1 \leq a_i \leq n

Each of the integers 1,...,n appears in the sequence.

n and a_i are integers.

Input

Input is given from Standard Input in the following format:

n
a_1 a_2 ... a_{n+1}

Output

Print n+1 lines.
The k-th line should contain the number of the different subsequences of the given sequence with length k, modulo 10^9+7.

Sample Input 1

3
1 2 1 3

Sample Output 1

3
5
4
1

There are three subsequences with length 1: 1 and 2 and 3.

There are five subsequences with length 2: 1,1 and 1,2 and 1,3 and 2,1 and 2,3.

There are four subsequences with length 3: 1,1,3 and 1,2,1 and 1,2,3 and 2,1,3.

There is one subsequence with length 4: 1,2,1,3.

Sample Input 2

1
1 1

Sample Output 2

1
1

There is one subsequence with length 1: 1.

There is one subsequence with length 2: 1,1.

Sample Input 3

32
29 19 7 10 26 32 27 4 11 20 2 8 16 23 5 14 6 12 17 22 18 30 28 24 15 1 25 3 13 21 19 31 9

Sample Output 3

32
525
5453
40919
237336
1107568
4272048
13884156
38567100
92561040
193536720
354817320
573166440
818809200
37158313
166803103
166803103
37158313
818809200
573166440
354817320
193536720
92561040
38567100
13884156
4272048
1107568
237336
40920
5456
528
33
1

Be sure to print the numbers modulo 10^9+7.


package main

import (
	"fmt"
)

func main() {
	var X, A, B int
	fmt.Scan(&X, &A, &B)
	if A-B >= 0 {
		fmt.Println("delicious")
	} else if B-A <= X {
		fmt.Println("safe")
	} else {
		fmt.Println("dangerous")
	}
}

Score : 100 points

Problem Statement

Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier.
He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though.

Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious.

Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it.

Write a program that outputs delicious if he found it delicious, safe if he did not found it delicious but did not get a stomachache either, and dangerous if he got a stomachache.

Constraints

1 ≤ X,A,B ≤ 10^9

Input

Input is given from Standard Input in the following format:

X A B

Output

Print delicious if Takahashi found the food delicious; print safe if he neither found it delicious nor got a stomachache; print dangerous if he got a stomachache.

Sample Input 1

4 3 6

Sample Output 1

safe

He ate the food three days after the "best-by" date. It was not delicious or harmful for him.

Sample Input 2

6 5 1

Sample Output 2

delicious

He ate the food by the "best-by" date. It was delicious for him.

Sample Input 3

3 7 12

Sample Output 3

dangerous

He ate the food five days after the "best-by" date. It was harmful for him.


package main

import (
	"fmt"
	"math"
)

func main() {
	var n, m int
	fmt.Scan(&n, &m)
	if AbsDiff(n, m) > 1 {
		fmt.Println(0)
		return
	}
	if n != m {
		fmt.Println(Factorial(n) * Factorial(m) % Mod)
	} else {
		fmt.Println(Factorial(n) * Factorial(m) * 2 % Mod)
	}
}

const Mod = 1000000007

func Min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

func Max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

func Factorial(n int) int {
	if n == 0 {
		return 0
	}
	num := 1
	for i := 2; i <= n; i++ {
		num = num * i % Mod
	}
	return num
}

func AbsDiff(n1 int, n2 int) int {
	return int(math.Abs(float64(n1 - n2)))
}

Score : 300 points

Problem Statement

Snuke has N dogs and M monkeys. He wants them to line up in a row.

As a Japanese saying goes, these dogs and monkeys are on bad terms. ("ken'en no naka", literally "the relationship of dogs and monkeys", means a relationship of mutual hatred.) Snuke is trying to reconsile them, by arranging the animals so that there are neither two adjacent dogs nor two adjacent monkeys.

How many such arrangements there are? Find the count modulo 10^9+7 (since animals cannot understand numbers larger than that).
Here, dogs and monkeys are both distinguishable. Also, two arrangements that result from reversing each other are distinguished.

Constraints

1 ≤ N,M ≤ 10^5

Input

Input is given from Standard Input in the following format:

N M

Output

Print the number of possible arrangements, modulo 10^9+7.

Sample Input 1

2 2

Sample Output 1

8

We will denote the dogs by A and B, and the monkeys by C and D. There are eight possible arrangements: ACBD, ADBC, BCAD, BDAC, CADB, CBDA, DACB and DBCA.

Sample Input 2

3 2

Sample Output 2

12

Sample Input 3

1 8

Sample Output 3

0

Sample Input 4

100000 100000

Sample Output 4

530123477


package main

import (
	"fmt"
	"math"
)

const MOD = int64(1e9 + 7)

func main() {
	var n, m int64
	fmt.Scan(&n, &m)

	if math.Abs(float64(n-m)) > 1 {
		fmt.Println(0)
		return
	}
	ans := (fac(n) * fac(m)) % MOD
	if n == m {
		ans = (ans * 2) % MOD
	}
	fmt.Println(ans)
}

func fac(num int64) int64 {
	if num == 1 {
		return 1
	}
	return (num * fac(num-1)) % MOD
}


package main

import (
	"strconv"
	"os"
	"bufio"
	"fmt"
)

func main() {
	var n int
	fmt.Scan(&n)
	a := make([]int, n);
	scanner := bufio.NewScanner(os.Stdin)
	scanner.Split(bufio.ScanWords)
	min, max := n, 0
	for i := 0; i < n; i++ {
		scanner.Scan()
		a[i], _ = strconv.Atoi(scanner.Text())
		if a[i] < min {
			min = a[i]
		}
		if a[i] > max {
			max = a[i]
		}
	}

	if max - min == 0 {
		if max == n - 1 || 2 * max <= n {
			fmt.Println("Yes")
		} else {
			fmt.Println("No")
		}
	} else if max - min == 1 {
		k := 0
		for i := 0; i < n; i++ {
			if a[i] == max - 1 {
				k++
			}
		}
		if max > k && 2 * (max - k) <= n - k {
			fmt.Println("Yes")
		} else {
			fmt.Println("No")
		}
	} else {
		fmt.Println("No")
	}
}

Score : 700 points

Problem Statement

There are N cats.
We number them from 1 through N.

Each of the cats wears a hat.
Cat i says: "there are exactly a_i different colors among the N - 1 hats worn by the cats except me."

Determine whether there exists a sequence of colors of the hats that is consistent with the remarks of the cats.

Constraints

2 ≤ N ≤ 10^5

1 ≤ a_i ≤ N-1

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print Yes if there exists a sequence of colors of the hats that is consistent with the remarks of the cats; print No otherwise.

Sample Input 1

3
1 2 2

Sample Output 1

Yes

For example, if cat 1, 2 and 3 wears red, blue and blue hats, respectively, it is consistent with the remarks of the cats.

Sample Input 2

3
1 1 2

Sample Output 2

No

From the remark of cat 1, we can see that cat 2 and 3 wear hats of the same color.
Also, from the remark of cat 2, we can see that cat 1 and 3 wear hats of the same color.
Therefore, cat 1 and 2 wear hats of the same color, which contradicts the remark of cat 3.

Sample Input 3

5
4 3 4 3 4

Sample Output 3

No

Sample Input 4

3
2 2 2

Sample Output 4

Yes

Sample Input 5

4
2 2 2 2

Sample Output 5

Yes

Sample Input 6

5
3 3 3 3 3

Sample Output 6

No


package main

import (
	"bufio"
	"fmt"
	"io/ioutil"
	"os"
)

type IOTp struct {
	iBuffer []byte
	iPos int
}
func (this *IOTp) GetChar() byte {
	res:=this.iBuffer[this.iPos]
	this.iPos++
	return res
}
func (this *IOTp) ReadInt() int {
	res:=0
	k:=1
	for ch:=this.iBuffer[this.iPos]; !(ch>='0' && ch<='9'); ch=this.iBuffer[this.iPos]{
		if ch=='-' {
			k=-1
		}
		this.iPos++
	}
	for ch:=this.iBuffer[this.iPos]; ch>='0' && ch<='9'; ch=this.iBuffer[this.iPos]{
		res=res*10+int(ch-'0')
		this.iPos++
	}
	res*=k
	return res
}
func NewIOTp() *IOTp {
	now:=new(IOTp)
	//inputFile,_:=os.Open("in.txt")
	inputFile:=os.Stdin
	now.iBuffer,_=ioutil.ReadAll(inputFile)
	now.iBuffer=append(now.iBuffer, '\n')
	now.iPos=0
	return now
}
var cin *bufio.Reader
var cout *bufio.Writer
const MAXN=100000
func main() {
	IO:=NewIOTp()
	cout=bufio.NewWriter(os.Stdout)
	n:=IO.ReadInt()
	var a,b [MAXN+10]int
	for i:=1; i<=n; i++ {
		a[i]=IO.ReadInt()
	}
	for i:=1; i<=n; i++ {
		b[i]=IO.ReadInt()
	}
	for _,v:=range a[1:n+1] {
		a[n+1]^=v
	}
	for _,v:=range b[1:n+1] {
		b[n+1]^=v
	}
	m:=make(map[int]int)
	for _,v:=range a[1:n+2] {
		m[v]++
	}
	for _,v:=range b[1:n+2] {
		m[v]--
	}
	for _,v:=range m {
		if v!=0 {
			fmt.Println(-1)
			return;
		}
	}
	ans:=0
	flag:=true
	for i:=range a[1:n+2] {
		if a[i]!=b[i] {
			if b[i]==a[n+1] {
				flag=false
			}
			ans++
		}
	}
	if a[n+1]==b[n+1] && ans!=0 && flag {
		ans++
	}
	fmt.Println(ans)
	cout.Flush()
}

Score : 1000 points

Problem Statement

There is a sequence of length N: a = (a_1, a_2, ..., a_N).
Here, each a_i is a non-negative integer.

Snuke can repeatedly perform the following operation:

Let the XOR of all the elements in a be x. Select an integer i (1 ≤ i ≤ N) and replace a_i with x.

Snuke's objective is to match a with another sequence b = (b_1, b_2, ..., b_N).
Here, each b_i is a non-negative integer.

Determine whether the objective is achievable, and find the minimum necessary number of operations if the answer is positive.

Constraints

2 ≤ N ≤ 10^5

a_i and b_i are integers.

0 ≤ a_i, b_i < 2^{30}

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N
b_1 b_2 ... b_N

Output

If the objective is achievable, print the minimum necessary number of operations.
Otherwise, print -1 instead.

Sample Input 1

3
0 1 2
3 1 0

Sample Output 1

2

At first, the XOR of all the elements of a is 3.
If we replace a_1 with 3, a becomes (3, 1, 2).

Now, the XOR of all the elements of a is 0.
If we replace a_3 with 0, a becomes (3, 1, 0), which matches b.

Sample Input 2

3
0 1 2
0 1 2

Sample Output 2

0

Sample Input 3

2
1 1
0 0

Sample Output 3

-1

Sample Input 4

4
0 1 2 3
1 0 3 2

Sample Output 4

5


package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
)

var sc = bufio.NewScanner(os.Stdin)

func Scan() string {
	sc.Scan()
	return sc.Text()
}

func iScan() int {
	n, _ := strconv.Atoi(Scan())
	return n
}

func fScan() float64 {
	n, _ := strconv.ParseFloat(Scan(), 64)
	return n
}
func stringToInt(s string) int {
	n, _ := strconv.Atoi(s)
	return n
}
func SScan(n int) []string {
	a := make([]string, n)
	for i := 0; i < n; i++ {
		a[i] = Scan()
	}
	return a
}
func iSScan(n int) []int {
	a := make([]int, n)
	for i := 0; i < n; i++ {
		a[i] = iScan()
	}
	return a
}
func abs(x int) int {
	if x < 0 {
		x = -x
	}
	return x
}
func larger(a, b int) int {
	if a < b {
		return b
	} else {
		return a
	}
}
func smaller(a, b int) int {
	if a > b {
		return b
	} else {
		return a
	}
}
func max(a []int) int {
	x := a[0]
	for i := 0; i < len(a); i++ {
		if x < a[i] {
			x = a[i]
		}
	}
	return x
}
func min(a []int) int {
	x := a[0]
	for i := 0; i < len(a); i++ {
		if x > a[i] {
			x = a[i]
		}
	}
	return x
}
func sum(a []int) int {
	x := 0
	for _, v := range a {
		x += v
	}
	return x
}

var mod int = 1000000007

func main() {
	buf := make([]byte, 0)
	sc.Buffer(buf, mod)
	sc.Split(bufio.ScanWords)
	n := iScan()
	a := iSScan(n)
	fmt.Println(max(a) - min(a))
}

Score : 200 points

Problem Statement

It is only six months until Christmas, and AtCoDeer the reindeer is now planning his travel to deliver gifts.

There are N houses along TopCoDeer street. The i-th house is located at coordinate a_i. He has decided to deliver gifts to all these houses.

Find the minimum distance to be traveled when AtCoDeer can start and end his travel at any positions.

Constraints

1 ≤ N ≤ 100

0 ≤ a_i ≤ 1000

a_i is an integer.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print the minimum distance to be traveled.

Sample Input 1

4
2 3 7 9

Sample Output 1

7

The travel distance of 7 can be achieved by starting at coordinate 9 and traveling straight to coordinate 2.

It is not possible to do with a travel distance of less than 7, and thus 7 is the minimum distance to be traveled.

Sample Input 2

8
3 1 4 1 5 9 2 6

Sample Output 2

8

There may be more than one house at a position.


package main

import (
	"bufio"
	"fmt"
	"io"
	"math"
	"os"
	"strconv"
)

var nextString func() string

func init() {
	nextString = newReadString(os.Stdin)
}

func main() {
	n := nextInt()
	mx := math.MinInt32
	mn := math.MaxInt32

	for i := 0; i < n; i++ {
		a := nextInt()
		mn = min(mn, a)
		mx = max(mx, a)
	}
	fmt.Println(mx - mn)
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

func nextInt() (result int) {
	result, _ = strconv.Atoi(nextString())
	return
}

func newReadString(ior io.Reader) func() string {
	r := bufio.NewScanner(ior)
	r.Buffer(make([]byte, 1024), int(1e+11))
	r.Split(bufio.ScanWords)

	return func() string {
		if !r.Scan() {
		}
		return r.Text()
	}
}