SCU 3578 H1N1's Problem

　　　　　　　　　　　　　　　　　　　　　　　　　　　3578: H1N1's Problem

Description

H1N1 like to solve acm problems.But they are very busy, one day they meet a problem.
Given three intergers a,b,c, the task is to compute a^(b^c))%317000011.
1412, ziyuan and qu317058542 don't have time to solve it, so the turn to you for help.

Input

The first line contains an integer T which stands for the number of test cases.
Each case consists of three integer a, b, c seperated by a space in a single line.
1 <= a,b,c <= 100000

Output

For each case, print a^(b^c)%317000011 in a single line.

Sample Input

2
1 1 1
2 2 2

Sample Output

1
16

快速幂+欧拉函数

地址 http://cstest.scu.edu.cn/soj/problem.action?id=3578

因为317000011为质数，所以对他求欧拉函数就是本身减一。直接两次快速幂，一开始类型没注意用longlong结果TLE了好几次，也是教训吧。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<queue>
#include<algorithm>
#include<map>
#include<iomanip>
#include<climits>
#include<string.h>
#include<numeric>
#include<cmath>
#include<stdlib.h>
#include<vector>
#include<stack>
#include<set>
#define INF 1e7
#define MAXN 100010
#define maxn 1000010
#define Mod 1000007
#define N 1010
using namespace std;
typedef long long LL;

LL quick_pow(LL a, LL n, LL mod)
{
    LL sum = 1;
    while (n)
    {
        if (n & 1) sum = (sum*a)%mod;
        a = (a%mod*a%mod) % mod;
        n >>= 1;
    }
    return sum%mod;
}    

void run()
{
    LL a, b, c, ans;
    scanf("%lld%lld%lld",&a,&b,&c);
    ans = quick_pow(a, quick_pow(b, c, 317000010), 317000011);
    printf("%lld
",ans);
}

int main()
{
    int T;
    scanf("%d", &T);
    while (T--)
        run();
    return 0;
}