SDUT 2605-A^X mod P(大幂分解求和)

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A^X mod P

Time Limit: 5000ms Memory limit: 65536K 有疑问？点这里^_^

题目描述

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.

示例输入

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

示例输出

Case #1: 14Case #2: 63

提示

来源

2013年山东省第四届ACM大学生程序设计竞赛

题意：求A^X mod P的和。

PS：妥妥的给跪了，用快速幂肯定不行，超时，然后用了一下快速幂+快速乘法还是超时，一直优化优化也优化好，只好去看了下题解，真心给跪了。

思路：这个用到了分解的方法，将A^f中的 f分解为 i * k + j的形式。保存在数组中，用的时候直接找就好了。

#include <stdio.h>#include <math.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <sstream>#include <algorithm>#include <set>#include <queue>#include <stack>#include <map>using namespace std;typedef long long LL;const int inf=0x3f3f3f3f;const double pi= acos(-1.0);const int maxn=33333;LL  X[maxn+10],Y[maxn+10];LL n,A,K,a,b,m,P;void Init(){    int i;    X[0]=1;    for(i=1;i<=maxn;i++){        X[i]=(X[i-1]*A)%P;    }    LL tmp=X[maxn];    Y[0]=1;    for(i=1;i<=maxn;i++){        Y[i]=(Y[i-1]*tmp)%P;    }}void Solve(int icase){    int i;    LL fx=K;    LL res=0;    for(i=1;i<=n;i++){        res=(res+(Y[fx/maxn]*X[fx%maxn])%P)%P;        fx=(a*fx+b)%m;    }   printf("Case #%d: %lld\n",icase,res);}int main(){    int T,icase;    scanf("%d",&T);    for(icase=1;icase<=T;icase++){        scanf("%lld %lld %lld %lld %lld %lld %lld",&n,&A,&K,&a,&b,&m,&P);        Init();        Solve(icase);    }}

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