Leftmost Digit（对数，数论）

Leftmost Digit

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

Total Submission(s): 14445 Accepted Submission(s): 5532

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

分析：

n^n=a.bcdegf.... * 10^m; 科学计数法

两边去对数：n*lg(n)=m+lg(a.bcdef...);（2）

因为在科学计数法中 1<=a<10, 所以 0<lg(a)<1；即，在（2）式中，lg(a.bcdef...)表示的是n*lg(n)的小数部分

code:

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<algorithm>
using namespace std;
int main()
{
    int t;
    long long n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld",&n);
        //n*lg(n)=m+lg(a);//1<=a<10,所以0<lg(a)<1 ，即表示的是小数部分
        //(int)n*lg(n)//表示的是m，
        // (double)n*lg(n)//表示的是 lg（a），
        double j=(double)n*log(n*1.0)/log(1.0*10);
        long long k=(long long)n*log(n)/log(10);
        int ans=(int)pow(10.0,j-k);
        cout<<ans<<endl;
    }
    return 0;

数学是深造的科学！！！