思路：刚开始做，发现很纠结！！
后来，看到了一个人的分析，是这样转换的
m=n^n;两边同取对数，得到，log10(m)=n*log10(n);再得到，m=10^(n*log10(n));
然后，对于10的整数次幂，第一位是1，所以，第一位数取决于n*log10(n)的小数部分
总之，log很强大啊，在求一个数的位数上，在将大整数化成范围内的整数上，在指数问题上
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).
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.
In the second case, 4 * 4 * 4 * 4 = 256, so the leftmost digit is 2.
代码：
#include<iostream>
#include<cmath>
using namespace std;
int main()
{
__int64 i,t,a,sum;
double s,x,n;
cin>>t;
while(t--)
{
cin>>n;
s=n*log10(n);
a=(__int64)s;
x=s-a;
sum=(__int64)pow(10.0,x);
cout<<sum<<endl;
}
return 0;
}