Rightmost Digit

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. 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。 View Code 1 #include<iostream> 2 using namespace std; 3 int main () 4 { 5 int test , n ; 6 int digit[25] = {0,1,4,7,6,5,6,3,6,9,0,1,6,3,6,5,6,7,4,9,0}; 7 cin>>test; 8 while (test -- ) 9 { 10 cin>>n; 11 cout<<digit[n%20]<<endl; 12 } 13 return 0 ; }