一个难题

题记

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

Output

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

Sample Input

7 66
8 800

Sample Output

9
6

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这个题简洁来说就是求a的b次方的最后一位数，因为是求最后一位数，所以可以通过取余10的方法来简化运算

同时还可以通过二次化简来使算法简洁化，也就是通过将b/2的方法来对式子简化，如果b是奇数，就将c与a相乘，得到新的c，之后将b除2，a自乘，以此构建一个循环，来求出最终结果的个位数c

	a %=10;
	while(b>0){
		if(b%2==1)
			c=c*a%10;
		b /=2;
		a = a*a %10;
	}
	return c;

综合代码：

#include 
int fei(int a,int b){
	int c=1;
	a %=10;
	while(b>0){
		if(b%2==1)
			c=c*a%10;
		b /=2;
		a = a*a %10;
	}
	return c;
} 
int main()
{
	int a,b,c;
	while(~scanf("%d %d",&a,&b)){
		c = fei(a,b);
		printf("%d\n",c);
	}
}