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


[align=left]Problem Description[/align]
Given a positive integer N, you should output the
leftmost digit of N^N.

[align=left]Input[/align]
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).

[align=left]Output[/align]
For each test case, you should output the leftmost
digit of N^N.

[align=left]Sample Input[/align]

2
3
4

[align=left]Sample Output[/align]

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.

[align=left]Author[/align]
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\n",s);
}
return 0;

}