HDOJ-----1097简单快速幂

Problem Description

lcy gives a hard puzzle to feng5166,lwg,JGShining and Ignatius: gave a and b,how to know the a^b.everybody objects to this BT problem,so lcy makes the problem easier than begin.
this puzzle describes that: gave a and b,how to know the a^b's the last digit number.But everybody is too lazy to slove this problem,so they remit to you who is wise.

 

Input

There are mutiple test cases. Each test cases consists of two numbers a and b(0<a,b<=2^30)

 

Output

For each test case, you should output the a^b's last digit number.

 

Sample Input

7 668 800

 

Sample Output

9
6
同余定理主要性质
（1）对于同一个除数，两数的和（或差）于他们余数的和（或差）同余数。（a + b)  %  c  =  (a % c + b % c) % c
（2）对于同一个除数，两数的乘积与他们余数的乘积同余。a * b % c = (a % c) * （b % c) % c
（3）对于同一个除数，如果两个整数同余，那么他们的差就一定能被这个数整除。a % c = b % c  则 (a - b) % c = 0
（4）对于同一个除数，如果两个整数同余，那么他们的乘方仍然同余。a % c = b % c 则 a^n % c = b^n % c
给定两个数a、b，求a的b次方最后一位数字，即a的b次方对10取余的快速幂
#include<cstdio>int main(){    int a, b, c;    while(~scanf("%d%d", &a, &b)){        int ans = 1;        a %= 10;        while(b > 0){            if(b & 1){                ans = (ans * a) % 10;            }            b >>= 1;            a = (a * a) % 10;        }        printf("%d\n", ans);    }    return 0;}