HDU 1061 Rightmost Digit
Rightmost Digit
Time
Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K
(Java/Others)
Total Submission(s): 12407 Accepted Submission(s):
4799
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).
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
可以发现一个规律：
当 n = 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 27 28 29 30 31 ...
rdigit = 1 4 7 6 5 6 3 6 9 0 1 6 3 6 5 6 7 4 9 0 1 4 7 6 5 6 3 6 9 0 ...
所以是以20为周期的规律。
1 #include<stdio.h>
2 int rdigit[25] = {0,1,4,7,6,5,6,3,6,9,0,1,6,3,6,5,6,7,4,9,0};
3 int main ()
4 {
5 int cas , n ;
6 scanf ( "%d" , &cas ) ;
7 while ( cas -- )
8 {
9 scanf ( "%d" , &n ) ;
10 printf ( "%d\n" , rdigit[n%20] ) ;
11 }
12 return 0 ;
13 }
