HDOJ--1097--A hard puzzle

A hard puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 34782    Accepted Submission(s): 12493

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

 

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

 题意：给两个数，求A^B的最后一位，跟HD--1061这个题大体上是一样的。

思路：计算幂然后进行取余就可以了，注意底数不能太大.(可以用取余操作底数)

ac代码：

#include
int  fun(int a,int b){
	int ans=1;
	a=a%10;
	while(b){
		if(b%2)
			ans=(ans*a)%10;
		a=(a*a)%10;
		b/=2;
	}
	return ans;
}
int main(){
	int a,b;
	while(scanf("%d%d",&a,&b)!=EOF){
		printf("%d\n",fun(a,b));
	}
	return 0;
}