poj 1995（快速幂）

题意：本题给出h组数每组两个数，ai，bi，求解 ;

分析：没啥好说的，一个裸的快速幂：

Source Code#include <set>#include <map>#include <stack>#include <queue>#include <math.h>#include <vector>#include <string>#include <utility>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <iostream>#include <algorithm>#include <functional>using namespace std;long long quick_mod(long long a,long long b,long long m){    long long ans=1;    while(b){        if(b&1){            ans=(ans*a)%m;            b--;        }        b/=2;        a=a*a%m;    }    return ans;}int main(){    int t;    scanf("%d",&t);    while(t--){        int a,b;        int m;        int h;        scanf("%d%d",&m,&h);        long long sum=0;        while(h--){             scanf("%d%d",&a,&b);             sum=(sum+quick_mod(a,b,m))%m;        }        printf("%I64d\n",sum%m);    }    return 0;}