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最新推荐文章于 2024-08-03 08:00:00 发布

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本文提供了一种高效方法来计算任意两个数相乘后末尾的数字，适用于0 < a, b <= 2^30的范围。通过简化计算过程，避免了直接乘法操作带来的复杂性。

A hard puzzle

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20787 Accepted Submission(s): 7317

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 66
8 800

Sample Output

9
6

Author

eddy

Recommend

JGShining

#include <iostream>
using namespace std;
int main()
{
	int a, b, n;
	while (cin>>a>>b) {
		a %= 10;
		b %= 4;
		switch(b) {
		case 1:
			n = a%10;
			break;
		case 2:
			n = a*a%10;
			break;
		case 3:
			n = a*a*a%10;
			break;
		case 0:
			n = a*a*a*a%10;
			break;
		default:
			break;
		}
		cout<<n<<endl;
	}
	return 0;
}