HDOJ---1061 Rightmost Digit[简单数学题]

最新推荐文章于 2025-08-22 13:45:39 发布

转载最新推荐文章于 2025-08-22 13:45:39 发布 · 67 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/XBWer/archive/2012/08/18/2645000.html

文章标签：

#java

本文介绍了一道编程题目RightmostDigit的解决方案，该题目要求输出正整数N的N次方最右侧的数字。通过使用预计算表格的方法，有效地解决了大规模输入情况下的计算效率问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19050 Accepted Submission(s): 7329

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. 

Author

Ignatius.L

code:

 1 #include<iostream>
 2 using namespace std;
 3 
 4 int table[10][10]={
 5     0,0,0,0,0,0,0,0,0,0,
 6     1,1,1,1,1,1,1,1,1,1,
 7     2,4,8,6,2,4,8,6,2,4,
 8     3,9,7,1,3,9,7,1,3,9,
 9     4,6,4,6,4,6,4,6,4,6,
10     5,5,5,5,5,5,5,5,5,5,
11     6,6,6,6,6,6,6,6,6,6,
12     7,9,3,1,7,9,3,1,7,9,
13     8,4,2,6,8,4,2,6,8,4,
14     9,1,9,1,9,1,9,1,9,1
15 };
16 
17 int cnt[10]={10,10,4,4,2,10,10,4,4,2};
18 
19 int main()
20 {
21     int t;
22     int n;
23     scanf("%d",&t);
24     while(t--)
25     {
26         scanf("%d",&n);
27         int temp=n;
28         temp%=10;
29         if(!temp)
30         {
31             printf("0\n");
32             continue;
33         }
34         printf("%d\n",table[temp][(n-1)%cnt[temp]]);
35     }
36     return 0;
37 }