HDUOJ 1060 Leftmost Digit（数学公式）

                                                Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15967    Accepted Submission(s): 6242

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

 

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题解：

计算出N^N最左边的数，就是最高位的数。

设N^N=d.xxxxx * 10^(k-1),其中k表示N^N的位数，那么d.xxxxx=10^lg(N^N-(k+1))，再对d.xxxx取整即可获得最终结果。因为k等于lgN^N的整数部分加一，即k=lgN^N+1（取整），所以d=10^(lgN^N-lg10N^N)(取整)。

AC代码：

#include<iostream>
#include<cmath>
using namespace std;
int main()
{
	int t,N;
	double x=0.0;
	cin>>t;
	while(t--){
		cin>>N;
		x=N*log10((double)N);
		x-=(long long)x;
		x=(int)pow(10,x);
		cout<<x<<endl;
	}
	return 0;
}