Rightmost Digit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 39389 Accepted Submission(s): 14863
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 6HintIn 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
#include <iostream> #include <algorithm> #include <cstdio> using namespace std; long long int powermod(long long int a,long long int b,long long int c) { long long int ans = 1; a = a%c; while (b > 0) { if (b% 2 ==1) ans = ans * a %c; b = b/2; a = a* a %c; } return ans; } int main() { int t; long long int n; cin >> t; while (t--) { cin >> n; cout <<powermod(n,n,10) << endl; } return 0; }
快速幂求余!