A^B mod C

Given A,B,C, You should quickly calculate the result of A^B mod C. (1<=A,B,C<2^63).

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 4

2 10 1000

Sample Output

1

24

我用的unsigned_int64开数据。用快速幂和快速乘写，注意：不要用%，会超时（大佬说的，并且我已经验证过了...233）

代码：

#include<stdio.h>

#define ull unsigned __int64

ull mul(ull a,ull b,ull c){

    ull res=0;

    a=a%c;

    while(b){

        if(b & 1)	res+=a;

        if(res>=c)	res-=c;	//代替%

        b/=2;

        a+=a;

        if(a>=c)  a-=c; 	//代替%

    }

    return res;

}

int main(){

	ull a,b,c,ans;

	while(scanf("%I64u%I64u%I64u",&a,&b,&c)!=EOF){

		ans=1;

		a=a%c;

		while(b){

			if(b%2==1)	ans=mul(ans,a,c);

			a=mul(a,a,c);

			b/=2;

		}

		printf("%I64u\n",ans);

	}

	return 0;

}

巴特西

A^B mod C (快速幂+快速乘+取模)题解

A^B mod C

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