快速幂取模

标签：leetcode

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


*/

快速幂取模

标签：leetcode

原文地址：http://blog.csdn.net/u014705854/article/details/46817083