Leftmost Digit

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 453 Accepted Submission(s): 229

Problem Description Given a positive integer N, you should output the leftmost 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 leftmost digit of N^N.

Sample Input 2 3 4

Sample Output 2 2 Hint In the first case, 3 * 3 * 3 = 27, so the leftmost digit is 2. In the second case, 4 * 4 * 4 * 4 = 256, so the leftmost digit is 2. #include <iostream> #include <stdio.h> #include <math.h> using namespace std; int main() { int num; double t,l,m; long long int N; cin>>num; while(num--) { cin>>N; t = N*log10(N); l = t - (long long int)t; m = pow(10,l); cout<<(int)m<<endl; } return 0; }