nefu 66 最左边的数

最左边的数
Problem:66 Time Limit:1000ms    Memory Limit:65536K

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.

Source

思路：设n^n=d.xxx*10^(k-1)，其中k为n^n的位数，所以k=(int)nlgn+1；

所以d.xxx=n^n / 10^(k-1)，分子可以转化为 10^lg(n^n)；

即d.xxx=10^nlgn / 10^(k-1)=10^[nlgn-(k-1)]=10^[nlgn-(int)nlgn]=pow(10,nlgn-(int)nlgn)；

最后取整，得d=(int)pow(10,nlgn-(int)nlgn)。

代码：

// 时隔好几个月，再来一发，以后要多写博客啊！！

#include <iostream>
#include <cmath>
using namespace std;


int main()
{
    int d,t,i;
    long long n;
    while(cin>>t)
    {
        for(i=1;i<=t;i++)
        {
            cin>>n;
            d=(int)(pow(10,(n*log10(n)-(int)(n*log10(n)))));
            cout<<d<<endl;
        }
    }
    return 0;
}