Hust oj 1521 Marshal's Confusion III(整数快速幂）_a ^(b ^ c)%317000011-优快云博客

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Marshal's Confusion III

Time Limit: 3000 MS

Memory Limit: 65536 K

Total Submit: 282(71 users)

Total Accepted: 89(60 users)

Rating:

Special Judge: No

Description

Marshallike 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. 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

Author

王勇

简单的整数快速幂，只不过从a^b mod c变成了a^b^c mod c，变形之后就是a^(b^c mod c-1)mod c,然后套模板

#include<iostream>
#include<cstring>

using namespace std;

const int mod=317000011;
long long  quickpow(long long   m , long long   n , long long   k)
{
    long long int ans = 1;
    while(n){
        if(n&1)
            ans = (ans * m) % k;
        n = n >> 1;
        m = (m * m) % k;
    }
    return ans;
}
int main()
{
    int n;
    long long int a,b,c;
    cin>>n;
    while(n--)
    {
        cin>>a>>b>>c;
        cout<<quickpow(a,quickpow(b,c,mod-1),mod)<<endl;;
    }
}