P1226-快速幂 

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 #define _for(i,a,b) for(int i = (a);i < b;i ++)
 4 typedef long long ll; 
 5 
 6 ll mi(ll a,ll b,ll m)
 7 {
 8     if(!b)
 9 
10         return (ll)0;
11 
12     ll rnt = 1;
13     while(b>0)
14 
15     {
16         if(b&0x1)
17 
18             rnt=(rnt*a)%m;
19         a = (a*a)%m;
20         b >>= 1;
21     }
22     return rnt;
23 }
24 int main()
25 {
26     ll a,b,m;
27 
28     scanf("%lld%lld%lld",&a,&b,&m);
29 
30     printf("%lld^%lld mod %lld=%lld\n",a,b,m,mi(a,b,m));
31 
32     return 0;
33 }

 

posted @ 2019-07-28 09:31  Asurudo  阅读(136)  评论(0)  编辑  收藏  举报