Rightmost Digit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22887 Accepted Submission(s): 8739
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
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 #include <cstdio> 2 #include <cstring> 3 #include <cmath> 4 #include <cstdlib> 5 using namespace std; 6 int main(void){ 7 #ifndef ONLINE_JUDGE 8 freopen("1061.in", "r", stdin); 9 #endif 10 int t; long long n; 11 while (~scanf("%d", &t)) 12 while (t--){ 13 scanf("%lld", &n); 14 int m = n % 10, l = n % 4 + 4, s = m; 15 for (int i = 1; i < l; ++i) s *= m; 16 printf("%d\n", s%10); 17 } 18 return 0; 19 }
这道题目还是很困惑，为什么说多test cases 还要加 while (~scanf("%d", &t)) 这句话？难道是每一个test case 又有多个cases?
因为这个原因卡了很久……次嗷……
