lintcode: Fast Power

Calculate the a^b % b where a, b and n are all 32bit integers.

Example For 231 % 3 = 2

For 1001000 % 1000 = 0

Challenge O(logn)

思路参见我的博客 幂模运算

class Solution {public:    /*     * @param a, b, n: 32bit integers     * @return: An integer     */    int fastPower(int a, int b, int n) {        // write your code here        long long res=1;//用int的话中间会溢出        long long base=a;        if(n==0){            return 1%b;        }        while(n){            int bi=n%2;            n/=2;            if(bi){                res=(res*base)%b;            }            base=(base*base)%b;        }        return res;    }};