hdu 1060 Leftmost Digit

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15032    Accepted Submission(s): 5813

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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和数学有关的题，要用到数学公式，不可以直接求出这个数。

m=n^n,两边分别对10取对数得 log10(m)=n*log10(n),得m=10^(n*log10(n)),由于10的任何整数次幂首位一定为1,所以m的首位只和n*log10(n)的小数部分有关

 

题意：输入N, 求N^N的首位是多少。

 

附上代码：

 

#include <iostream>
#include <cmath>
#include <cstdio>
using namespace std;
int main()
{
    __int64 t;
    int n,m,s;
    double a,b;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%d",&m);
        a=1.0*m*log10(m*1.0);
        t=(__int64)a;
        b=a-t;           //求出小数部分
        s=(int)(pow(10,b));
        printf("%d
",s);
    }
    return 0;

}