A^X mod P 大幂分解求和打表

心已赠人 2022-06-16 00:07 120阅读 0赞

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  
  
  
Sample Output  
  
Case \#1: 14  
Case \#2: 63

#include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <queue>
    #include <cmath>
    #include <algorithm>
    #include <vector>
    #include <map>
    #include <string>
    #include <stack>
    #define LL long long
    #define INF 0x7fffffff
    #define maxn 33333
    #define PI 3.1415926535897932
    #define E 2.718281828459045
    using namespace std;
    long long t,n,A,K,a,b,m,p;
    long long X[maxn+10],Y[maxn+10];   //x[maxn]里面存着A^(1到maxn) y[maxn]里面存着 A^(maxn=10000000000多）次方  将表拆开进行记录
    void init()  //打表；
    {
        int i;
        X[0]=1;
        for(i=1;i<=maxn;i++){
            X[i]=(X[i-1]*A)%p;
        }
        Y[0]=1;
        long long ans=X[maxn];
        for(i=1;i<=maxn;i++){
            Y[i]=(Y[i-1]*ans)%p;
        }
    }
    int main()
    {
        int cases=1;
        scanf("%d",&t);
        while(t--)
        {
            scanf("%I64d%I64d%I64d%I64d%I64d%I64d%I64d",&n,&A,&K,&a,&b,&m,&p);
            init();
            long long sum=0;
            for(long long i=1;i<=n;i++){
                sum=(sum+(X[K%maxn]*Y[K/maxn])%p)%p;  //把K分解
                K=(a*K+b)%m;
            }
            cout<<"Case #"<<cases<<": "<<sum<<endl;
            cases++;
        }
        return 0;
    }