Rightmost Digit
2017-07-18 10:46
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Rightmost Digit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 56655 Accepted Submission(s): 21400
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.
解题思路:本题有规律可寻个位数相乘mod10会出现循环
if (ans == 0 || ans == 1 || ans == 5 || ans == 6) return ans; if (ans == 2 || ans == 3 || ans == 7 || ans == 8) // T = 4 if (ans == 4 || ans == 9) // T = 2
具体实现
#include <stdio.h> int main() { int num; scanf("%d",&num); while(num--){ long long value; scanf("%lld",&value); int ans = value % 10; if (ans == 0 || ans == 1 || ans == 5 || ans == 6) printf("%d\n",ans); else { int a[10][10] = {0}; for (int i = 2; i < 10; ++i){ int k = 1; for (int j = 1; j < 10; ++j){ k *= i; a[i][j] = k%10; } } if (ans == 4 || ans == 9) printf("%d\n",a[ans][value % 2 + 2]); else printf("%d\n",a[ans][value % 4 + 4]); } } return 0; }
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