HDU1060 Leftmost Digit 【数学】

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 12897    Accepted Submission(s): 4934

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. 
    
    
    
    
     
   
   
   
   

 

因为N^N = A * 10^X (科学计数法,A即为所求最高位)，两边同时取对数得N * log10(N) = log10(A) + X, 得 log10(A) = N * log10(N) - X, 得 A = 10^(N*log10(N) - X), 其中X为N^N结果的位数-1，即[log10(N*N)], 中括号表示取整，所以最终结果为A = 10^(N*log10(N) - [N*log10(N)]).需要注意的是N^N的位数要用__int64类型存储。

#include <stdio.h>
#include <math.h>

int main()
{
    int t, n;
    double ans;
    scanf("%d", &t);
    while(t--){
        scanf("%d", &n);
        ans = n * log10(n);
        ans -= (__int64)ans;
        ans = pow(10, ans);
        printf("%d\n", (int)ans);
    }
    return 0;
}