HDOJ 1060 Leftmost Digit

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16383    Accepted Submission(s): 6481

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

 

这个题我记得我之前做过啊，结果博客上并没有。因为这个题还是挺有用的所以现在写一下贴上。

题解：

这个数为num的话，num^num=a*10ⁿ

两边同时lg → num * lg(num) = lg(a) + n

lg(a)肯定是小数部分，所以把整数部分减掉得到lg(a)

最后把a还原，求出第一位。

#include<stdio.h>
#include<math.h>
int main()
{
	int T;
	int i,n;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d",&n);
		double x=n*log10(1.0*n);
		x-=(__int64)x;
		int ld=pow(10,x);
		printf("%d\n",ld);
	}
	return 0;
}