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1.41 kB
| using namespace std; | |
| long long exp(long long a, long long b, long long p){ | |
| long long res = 1; | |
| while(b){ | |
| if(b&1) res = (res*a)%p; | |
| a = (a*a)%p; | |
| b >>= 1; | |
| } | |
| return res; | |
| } | |
| /*Divide and conquer approach for finding the sum of a GP in O(log^2 n) time*/ | |
| long long gp(long long r, long long n, long long p){ | |
| if(n == 0) | |
| return (1%p); | |
| else if(n == 1) | |
| return (1+r)%p; | |
| else{ | |
| if(n&1){ | |
| return ( ( (1 + exp(r,(n+1)/2,p) )%p) * gp(r,(n-1)/2,p) )%p; | |
| } | |
| else{ | |
| return (((( (1 + exp(r,n/2,p)) %p) * gp(r,(n-1)/2,p) )%p) + exp(r,n,p))%p; | |
| } | |
| } | |
| } | |
| int main(){ | |
| int t; | |
| int a,d,r,n,MOD; | |
| long long x,sum; | |
| sd(t); | |
| while(t--){ | |
| scanf("%d %d %d",&a,&d,&r); | |
| scanf("%d %d",&n,&MOD); | |
| x = (a * exp(r,(n-1)/2,MOD))%MOD; | |
| sum = (d * gp(r,(n-1)/2,MOD) )%MOD; | |
| if(n&1){ | |
| printf("%lld\n",(((x+sum)%MOD) - (d%MOD) + MOD)%MOD); | |
| } | |
| else{ | |
| printf("%lld\n",(x+sum)%MOD); | |
| } | |
| } | |
| return 0; | |
| } | |