Hust oj 1521 Marshal's Confusion III(整数快速幂）

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Marshal's Confusion IIITime Limit: 3000 MSMemory Limit: 65536 KTotal Submit: 282(71 users)Total Accepted: 89(60 users)Rating:

Special Judge: NoDescriptionMarshallike 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. InputThe 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 <= 100000OutputFor each case, print a^(b^c)%317000011 in a single line.Sample Input2
1 1 1
2 2 2Sample 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;;    }}

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