hdu1061 Rightmost Digit
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Rightmost Digit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 29900 Accepted Submission(s): 11382
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
234
Sample Output
76HintIn 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.
//============================================================================// Name : Math_hdu1061.cpp// Author : vit// Version :// Copyright : Your copyright notice// Description : Hello World in C++, Ansi-style//============================================================================#include <iostream>using namespace std;int nu[10] = { 1, 1, 4, 4, 2, 1, 1, 4, 4, 2 };//找规律int main() {int t;long long a, temp, b;cin >> t;while (t--) {cin >> temp;a = temp % 10;if (nu[a] == 1) {cout << a << endl;continue;}else if (a == 4) {cout << '6' << endl;continue;}else if (a == 9){cout << '9' << endl;continue;}b = temp % nu[a];if(a == 2){switch(b){case 0:cout << '6' << endl;break;case 1:cout << '2' << endl;break;case 2:cout << '4' << endl;break;case 3:cout << '8' << endl;break;default:break;}}else if(a == 3){switch(b){case 0:cout << '1' << endl;break;case 1:cout << '3' << endl;break;case 2:cout << '9' << endl;break;case 3:cout << '7' << endl;break;default :break;}}else if(a == 8){switch(b){case 0:cout << '6' << endl;break;case 1:cout << '8' << endl;break;case 2:cout << '4' << endl;break;case 3:cout << '2' << endl;break;default:break;}}else if(a == 7){switch(b){case 0:cout << '1' << endl;break;case 1:cout << '7' << endl;break;case 2:cout << '9' << endl;break;case 3:cout << '3' << endl;break;default:break;}}}return 0;}
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