快速幂取模模板

Posted 2020-08-29 R o b i n

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int MOD;
 4 
 5 int fast_pow_mod(int a, int b) {
 6     int res = 1;
 7     while(b) {
 8         if (b & 1) res = res * a % MOD;
 9         a = a * a % MOD;
10         b >>= 1;
11     }
12     return res;
13 }
14 
15 int main() {
16     int a, b;
17     cin>>a>>b>>MOD;
18     cout<<fast_pow_mod(a, b)<<endl;
19     return 0;
20 }