Leftmost Digit

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

Total Submission(s): 15305    Accepted Submission(s): 5937

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

 

Solution

取对数。

设 n^n = x*10^y（n^n的科学计数法表示），

则 log (n^n) = y + log x

 

Implementation

#include 
      
       using namespace std;int main(){    int T;    scanf("%d", &T);    for(int n; T--; ){        scanf("%d", &n);        double t=log10(n)*n;        printf("%d\n",(int)pow(10, t-floor(t)));    }}
      

 

Tips

<cmath>内常用函数：

pow 

exp

log　　自然对数

log10　常用对数

log2　　以2为底的对数

floor

ceil