hdu-oj 1061 Rightmost Digit_hduoj1061-优快云博客

Rightmost Digit

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

Total Submission(s): 31706 Accepted Submission(s): 12130

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

Hint

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
题目大意：输出N^N.的个位数字
解题思路： 20一个周期有规律
附代码：
#include<iostream>   
using namespace std;   
int main()   
{   
    int  rdigit[20] = {0,1,4,7,6,5,6,3,6,9,0,1,6,3,6,5,6,7,4,9};   
    int n;   
    cin>>n;
	while(n--)
	{
            int temp;   
               cin>>temp;   
	  cout<<rdigit[temp%20]<<endl;   
        }
    
    return 0;   
}
计算机找规律代码：
#include<stdio.h>
int main()
{
	long int n;
	int i,a[100],j,t,y;
	scanf("%d",&t);
	for(i=0;i<t;i++)
	{
		scanf("%d",&n);
		y=n%10;a[0]=y;
		for(j=1;;j++)//找规律周期
		{
			a[j]=a[j-1]*y;
			if(a[j]%10==y)
				break;
		}
		n=n%j;
		if(n==0)
			n=n+j;
		a[n-1]=a[n-1]%10;
		printf("%d\n",a[n-1]);
	}
}