HDU 1060:Leftmost Digit【数学】

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15834    Accepted Submission(s): 6177

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. 
    
    
    
    
     思路：
    
    
    
    
把num*num的值记作：num*num=a*10^n，其中1<a<10;
那么，通过两边取对数的方法得到num*log10(1.0*num)=log10(a)+n,这时0<log10(a)<1;
令x=n+log10(a),得到log10(a)=x-n;所以a=10^(x-n)；
n为整数部分，log10(a)为小数部分，由x=n+log10(a),可知(int)x=n；
最终a=10^（x-n）=10^(x-(int)x)！
    
    
    
    
AC-code：
    
    
    
    
#include<cstdio>
#include<cmath>
int main()
{
	long long n;
	int T;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%lld",&n);
		double x=n*log10(n*1.0);
		double a=x-(long long)x;
		int ans=pow(10.0,a);
		printf("%d\n",ans);
	}
 }