A^X mod P（山东第四届省赛）

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

2

3 2 1 1 1 100 100

3 15 123 2 3 1000 107

Output

Case #1: 14
Case #2: 63

Hint

 

Source

2013年山东省第四届ACM大学生程序设计竞赛

 

 

附上一段会T的代码：

#include <stdio.h>
#include <vector>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
#define ll long long
const int inf = 0xffffff;
const int maxn = 1e6+8;
int n, t;
ll A, K, a, b, m, P, f[maxn], fi[maxn];
ll qsm(ll x, ll y, ll z)
{
    ll ans = 1;
    while(y)
    {
        if(y&1)
        {
            ans = ans*x%z;
        }
        x = x*x%z;
        y >>= 1;
    }
    return ans;
}
int main()
{
    scanf("%d", &t);
    int ga = t;
    while(t--)
    {
        ll sum;
        scanf("%d%lld%lld%lld%lld%lld%lld", &n, &A, &K, &a, &b, &m, &P);
        ll miao = K;//存下第一项，f(x) = k
        sum = qsm(A, K, P);//存下第一项 A^(f(1))%P
        for(int i = 1; i<n; i++)
        {
            miao = (a*miao+b)%m;//由题可知，后一项等于 a*前一项+b，这里是依次算出以后的每一个f(i)
            sum = (sum+qsm(A, miao, P))%P;//这里是依次算出A^(f(i))%P
        }
        printf("Case #%d: %lld
", ga-t, sum);
    }
    return 0;
}