快速幂取模

用二分，使复杂度由 O(n) 变为 O(logn)

#include<iostream>#include<cstdio>using namespace std;/// (b^n)mod m;  (a*b mod m) = (a mod m)*(b mod m)mod m       O(log n)/// (b^n)mod m;  (a - b ) mod m =( (a mod m)-(b mod m) + n ) mod m  )      O(log n)/// (b^n)mod m;  (a + b ) mod m =( (a mod m)+(b mod m) mod m  )       O(log n)int mod(int b,int n,int m) // 位运算{    int ans=1;    while(n){        if(n&1)            ans=(ans*b)%m;        b=(b*b)%m;        n=n>>1;    }    return ans;}int mod_1(int b,int n,int m) //  O(log n) 递归{    if(n==0) return 1;    int x=mod_1(b,n/2,m);    long long ans=(long long)x*x%m;    if(n%2) ans=ans*b%m;   /// n&1    return (int)ans;}int main(){    int b,n,m;    while(scanf("%d%d%d",&b,&n,&m))    {        printf("%d  ",mod(b,n,m));        printf("%d\n",mod_1(b,n,m));    }    return 0;}/**12996 227 37909  == 7775*/