< 返回

# Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22791    Accepted Submission(s): 8697 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.
1. //打表法可以发现一个规律：
2. //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 ...
3. //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    ...
4. //所以是以20为周期的规律。
5. #include <iostream>
6. using namespace std;
7. int main()
8. {
9. int s = {0, 1, 4, 7, 6, 5, 6, 3, 6, 9, 0, 1, 6, 3, 6, 5, 6, 7, 4, 9, 0} ;
10.  int t, n;
11.     cin >> t;
12. while (t --)
13. {
14.    cin >> n;
15.    cout << s[n % 20] << endl;
16. }
17. return 0;
18. }

/template/Home/DawnNew/PC/Static