（大幂分解求和）A^X mod P(数的高次幂）

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. 

输入

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

输出

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.

样例输入 Copy

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

样例输出 Copy

Case #1: 14
Case #2: 63


就是预先把A的次幂处理出来

dp1数组构造A^0~A^(10^5)，间隔为A。

dp2数组构造A^(10^5)~A^(10^10)，间隔为A^(10^5)。

这样对于任意的A^x就能表示成dp2[x/(10^5)]*dp1[x%(10^5)]

从而用空间换取时间。

如果是A的1e4次幂，表示为dp2[0]*dp1[1e4]=dp1[1e4]

如果是A的1e6次幂，表示为dp2[10]*dp1[0]

如果是A的1e6+10次幂，表示为dp2[10]*dp[10]=A的1e6次幂乘A的10次幂，就是A的1e6+10次幂

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e5;
ll f1[110011];
ll f2[110011];
ll n,A,k,a,b,m,p;
void inint(){
    f1[0]=f2[0]=1;
    f1[1]=A%p;
    for(int i=2;i<=maxn;i++){
        f1[i]=(f1[i-1]*f1[1])%p;
    }
    f2[1]=f1[maxn];
    for(int i=2;i<=maxn;i++){
        f2[i]=(f2[i-1]*f2[1])%p;
    }
} 
int main(){
    int t;
    cin>>t;
    for(int case1=1;case1<=t;case1++){
        cin>>n>>A>>k>>a>>b>>m>>p;
        inint();
        ll ans=0;
        ll z=k;
        for(int i=1;i<=n;i++){
            ans=(ans+f2[z/maxn]*f1[z%maxn])%p;
            z=(a*z+b)%m;
        }
        printf("Case #%d: %lld
",case1,ans);
    }
}