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;
}