A^B mod C (快速幂+快速乘+取模)题解

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;
}