hdoj 1061 Rightmost Digit 【数论】

两次TL 伤不起的找规律啊

来源

http://blog.csdn.net/litiouslove/article/details/7823363

参考代码：

/*末位数      相乘后的末位数1           12           4   8   6   23           9   7   1   34           6   45           56           67           9   3   1   78           4   2   6   89           1   9（由于x9是奇数，所以1不会出现）0           0由上面的分析可见，每个数相乘后最多有四个结果所以对一个数n，只需做其对4取余后余数次相乘即可但是会出现 n%4 == 0 的情况由于x5^x5的末位数是5，x9^x9的末位数是9所以将n%4转换为(n-1)%4*/

 

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 31298    Accepted Submission(s): 11939

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

234

 

Sample Output

76
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.
 

 

代码：

#include <stdio.h>int main(){ int n, t; scanf( "%d", &t ); while( t -- ){  scanf( "%d", &n );  int temp = n%10;  int ans = 1;  for( int i = 0; i <= (n-1)%4; i ++ )  ans = ans*temp;  printf( "%d\n", ans%10 ); } return 0;}