杭电ACM 1061

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 47234    Accepted Submission(s): 17809

Problem Description

Given a positive integer N, you should output the most right digit of N^N.

 

Input

The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains a single positive integer N(1<=N<=1,000,000,000).

 

Output

For each test case, you should output the rightmost digit of N^N.

 

Sample Input

2
3
4

 

Sample Output

 

7

6

In the first case, 3 * 3 * 3 = 27, so the rightmost digit is 7.
In the second case, 4 * 4 * 4 * 4 = 256, so the rightmost digit is 6.

解题思路：由于题目要求只需要输出个位数即可，我们便可以从输入数的个位数着手1,2,3.......我们可以以下分两类。

第一类：0,1,5,6它们四个无论是多少次方它的个位都不会变

第二类：例如个位数是2它的n次方后的个位可能的情况是2,4,8,6,2,4,8,6。题目要求是N的N次方，所以奇数次方是取不到的，只能取4,6。这是偶数的情形，奇数也一样。不难发现4是它们的周期。到了这里，你只需要分析它余4是几判断下就好了。

#include<iostream>
using namespace std;
int main()
{
	int n,t;
	cin>>t;
	while(t--)
	{
		cin>>n;
		if(n % 10 == 0 || n % 10 == 1 || n % 10 == 5 || n % 10 == 6 ||  n % 10 == 9)  cout<<n % 10<<endl;
		else if(n % 10 == 4)  cout<<6<<endl;
		else if(n % 10 == 2)
		{
			if(n %4 == 2)  cout<<4<<endl;
			else  cout<<6<<endl;
		}
		else if(n % 10 == 3)
		{
			if(n %4 == 1)  cout<<3<<endl;
			else cout<<7<<endl;
		}
		else if(n % 10 == 7)
		{
			if(n %4 == 1)  cout<<7<<endl;
			else cout<<3<<endl;
		}
		else if(n % 10 == 8)
		{
			if(n %4 == 2)  cout<<4<<endl;
			else cout<<6<<endl;
		}		
	}
    return 0;
}