HDOJ 1060 Leftmost Digit(数学，求n^n的最高位)



Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15428    Accepted Submission(s): 5983

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.  
    
    
    
    
     
   
   
   
   

 

题意：求n^n的最高位的数字

题解：数学渣哭一会，推导能力太差啊

首先我们用科学计数法可以表示 num^num = a * 10^(n)  （a就是最高位数字）

等式两边同时用log10取对数得  num * lg(num) = n + lg(a)

设 x = n + lg(a) = num * lg(num)

因为a小于10，所以n为x的整数部分， lg(a)为x的小数部分；

又因为lg(a) = x-n， 所以lg(a) = x - (int)x  得到： a = pow( 10.0 , x-(int)x ) 。

又因为x = num * lg(num) ，所以就能得到 n^n的最高位的数字。

代码如下：

#include<cstdio>
#include<cmath>
#include<cstring>
int main()
{
	int t,n;
	scanf("%d",&t);
	while(t--)
	{
		__int64 n;
		scanf("%I64d",&n);
		double x=n*log10(n*1.0);
		x-=(__int64)x;
		int ans=pow(10.0,x);
		printf("%d\n",ans);
	}
	return 0;
}