快速幂求a^b结果的最后一位

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

96

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;


int k_mi(int a, int b) { //快速幂取模 
int ans = 1,t = a % 10;
while(b) {
if(b&1) ans = (ans * t) % 10;
t = (t * t) % 10;
b >>= 1;
}
return ans % 10;
}


int main() {
int a, b;
while(scanf("%d %d", &a, &b) != EOF) {
printf("%d\n", k_mi(a, b));
}
return 0;
}