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hdu-1061 Rightmost Digit

2024-09-30 22:54:39   211人阅读

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]);     }}

 

hdu-1061 Rightmost Digit


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