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| // Connect the Dots | |
| // Solution by Jacob Plachta | |
| using namespace std; | |
| template<typename T> T Abs(T x) { return(x<0 ? -x : x); } | |
| template<typename T> T Sqr(T x) { return(x*x); } | |
| string plural(string s) { return(Sz(s) && s[Sz(s)-1]=='x' ? s+"en" : s+"s"); } | |
| const int INF = (int)1e9; | |
| const LD EPS = 1e-12; | |
| const LD PI = acos(-1.0); | |
| bool Read(int &x) | |
| { | |
| char c,r=0,n=0; | |
| x=0; | |
| for(;;) | |
| { | |
| c=GETCHAR(); | |
| if ((c<0) && (!r)) | |
| return(0); | |
| if ((c=='-') && (!r)) | |
| n=1; | |
| else | |
| if ((c>='0') && (c<='9')) | |
| x=x*10+c-'0',r=1; | |
| else | |
| if (r) | |
| break; | |
| } | |
| if (n) | |
| x=-x; | |
| return(1); | |
| } | |
| int main() | |
| { | |
| if (DEBUG) | |
| freopen("in.txt","r",stdin); | |
| // vars | |
| int T,t; | |
| int N,H,V; | |
| int i,j,si,a,b,c,d; | |
| static int G[2][LIM]; | |
| static PR P[LIM]; | |
| static int maxY[LIM]; | |
| set<PR> S; | |
| set<PR>::iterator I; | |
| // testcase loop | |
| Read(T); | |
| Fox1(t,T) | |
| { | |
| // input, and generate coordinates | |
| Read(N),Read(H),Read(V); | |
| Fox(i,2) | |
| { | |
| Read(G[i][0]),Read(G[i][1]),Read(a),Read(b),Read(c),Read(d); | |
| FoxI(j,2,N-1) | |
| G[i][j]=((LL)a*G[i][j-2] + (LL)b*G[i][j-1] + c)%d+1; | |
| } | |
| // impossible? | |
| int ans=-1; | |
| if (H+V<N) | |
| goto Done; | |
| // sort dots by X | |
| Fox(i,N) | |
| P[i]=mp(G[0][i],G[1][i]); | |
| sort(P,P+N); | |
| // precompute Y values' suffix maxes | |
| maxY[N]=0; | |
| FoxR(i,N) | |
| maxY[i]=max(maxY[i+1],P[i].y); | |
| // consider each possible max horiz. segment length | |
| ans=2*INF; | |
| S.clear(); | |
| S.insert(mp(0,-1)); | |
| I=S.begin(),si=0; | |
| Fox(i,N+1) | |
| { | |
| // update set of dots so far | |
| if (i) | |
| { | |
| S.insert(mp(P[i-1].y,i)); | |
| if (mp(P[i-1].y,i)<*I) | |
| si++; | |
| } | |
| // last f dots must use vert. segments | |
| int f=N-i; | |
| if (f>V) | |
| continue; | |
| // g of the dots so far must also use vert. segments | |
| int g=max(0,min(N-H,V)-f); | |
| while (si>g) | |
| I--,si--; | |
| while (si<g) | |
| I++,si++; | |
| Min(ans,(i?P[i-1].x:0)+max(I->x,maxY[i])); | |
| } | |
| // output | |
| Done:; | |
| printf("Case #%d: %d\n",t,ans); | |
| } | |
| return(0); | |
| } |