杭电 HDU 1060 Leftmost Digit

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14198    Accepted Submission(s): 5437

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.

 

Author

Ignatius.L

 #include<iostream>
#include<cmath>
using namespace std;
int main()
{
  int  n;int T;
     cin>>T;
  while(T--)
  {
   scanf("%d",&n);
  double sum1=n*log10(double(n));
   double sum2=sum1-floor(sum1);
   printf("%d\n",int (pow(10,sum2)));
  }
  return 0;
}