C

Description

It's easy for ACMer to calculate A^X mod P. Now given seven integers n, A, K, a, b, m, P, and a function f(x) which defined as following.

f(x) = K, x = 1

f(x) = (a*f(x-1) + b)%m , x > 1

Now, Your task is to calculate

( A^(f(1)) + A^(f(2)) + A^(f(3)) + ...... + A^(f(n)) ) modular P. 

Input

  In the first line there is an integer T (1 < T <= 40), which indicates the number of test cases, and then T test cases follow. A test case contains seven integers n, A, K, a, b, m, P in one line.

1 <= n <= 10^6

0 <= A, K, a, b <= 10^9

1 <= m, P <= 10^9

Output

  For each case, the output format is “Case #c: ans”. 

c is the case number start from 1.

ans is the answer of this problem.

Sample Input

23 2 1 1 1 100 1003 15 123 2 3 1000 107

Sample Output

Case #1: 14

Case #2: 63

思路：

f(x)分解成f(x)=q*x+j;所以A^f(x)=(A^x)^q*A^j;

即将f(x)的值分段存，即A^f(x)分段存（将A^f(x)分为每段长为33333的33333段，求A^f(x)时先求在哪一段里面，再求这一段里面具体哪个位置。

#include<iostream>#include<string.h>#include<algorithm>using namespace std;#define maxn 33333long long x[2*maxn],y[2*maxn];long long n,A,K,a,b,m,p;void init(){    x[0]=1;    for(int i=1;i<=maxn;i++){        x[i]=(x[i-1]*A)%p;    }    long long temp=x[maxn];    y[0]=1;    for(int i=1;i<=maxn;i++){        y[i]=(y[i-1]*temp)%p;    }}void solve(int tx){    long long fx=K;    long long ans=0;    for(int i=1;i<=n;i++){        ans=(ans+(y[fx/maxn]*x[fx%maxn])%p)%p;        fx=(a*fx+b)%m;    }    cout<<"Case #"<<tx<<": "<<ans<<endl;}int main (){    int t;    cin>>t;    for(int l=1;l<=t;l++){        cin>>n>>A>>K>>a>>b>>m>>p;        init();        solve(l);    }}