hdu1060

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5699    Accepted Submission(s): 2166

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

234

 

 

Sample Output

22
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.
 

 

 

m=n^n,两边分别对10取对数得 log10(m)=n*log10(n),得m=10^(n*log10(n)),由于10的任何整数次幂首位一定为1,所以m的首位只和n*log10(n)的小数部分有关

 #include<stdio.h>
#include<math.h>
int main()
{
 __int64 n;
 int t;
 scanf("%d",&t);
 while(t--)
 {
  scanf("%I64d",&n);
  double ans1=n*log10(double(n));
  __int64 ans2=__int64(ans1);
  double ans=ans1-ans2;
  int left_digit=int(pow(10.0,ans));
  printf("%d\n",left_digit);
 }
 return 0;
}