hdu-1061 Rightmost Digit

Posted 2020-09-21

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=1061

题目类型：

水题

题意概括：

求n的n次方的个位数。

解题思路：

因为N的范围太大，所以我通过对位数为1-9的数进行20次次方打表，发现他们的循环节不是4，就是4的因子，所以我对位数为1-9进行打表四次，然后对输入的数只判断个位数，然后判断这个数在第几循环节并输出即可。

题目：

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 55914    Accepted Submission(s): 21129

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.

 

# include <stdio.h>
# include <string.h>
int main ()
{
    int i,j,t,n,x,y;
    int a[10][10];
    memset(a,0,sizeof(a));
    for(i=1;i<10;i++)
    {
        a[i][1]=i;
        for(j=2;j<5;j++)
        {
            a[i][j]=a[i][j-1]*i%10;
        }
        a[i][0]=a[i][4];
     }
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d",&n);
         x=n%10;
         y=n%4;
         printf("%d\n",a[x][y]);
     }
}