AcWing每日一题--a^b

https://www.acwing.com/problem/content/91/

考察快速幂的知识;

 1 #include<iostream>
 2 using namespace std;
 3 typedef long long LL;
 4 LL ksm(LL a,LL b,LL p){
 5     LL res=1;
 6     while(b){
 7         if(b&1){
 8             res=(res*a)%p;
 9         }
10         a=a*a%p;
11         b>>=1;
12     }
13     return res%p;
14 }
15 int main(void){
16     LL a,b,p;
17     cin>>a>>b>>p;
18     cout<<ksm(a,b,p);
19     return 0;
20 }
原文地址:https://www.cnblogs.com/greenofyu/p/14382592.html