A^B mod C

Problem Description

Given A,B,C, You should quickly calculate the result of A^B mod C. (1<=A,B,C<=1000000000).

 Input

There are multiply testcases. Each testcase, there is one line contains three integers A, B and C, separated by a single space.

 Output

For each testcase, output an integer, denotes the result of A^B mod C.

 Sample Input

3 2 42 10 1000

 Sample Output

124

#include<stdio.h>__int64 power(__int64 a,__int64 b,__int64 c){__int64 ans=1;while(b){if(b&1)  //当b为奇数时{ans=(ans*a)%c;   // a的99次方的等于a的49次方的平方，还要再乘以ab--;}b/=2;   //二分a=a*a%c;    //  a的98次方的等于a的49次方的平方}return ans;}int main(){long  a,b,c;while(scanf("%ld%ld%ld",&a,&b,&c)!=EOF){printf("%ld\n", power(a,b,c));}return 0;}

采用了二分法快速幂