hdu1060

之前做过的一道题，现在还是拿出来谢谢自己的心得，不是为了叠加博客的内容，而是为了这样每做一道题会一种方法，坚持下去（多吃饭，多运动，多敲代码，多写博客，多陪女朋友，）为了自己的健康，自己开心快乐

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.

代码

#include<iostream>
#include<cmath>
using namespace std;
int main()
{
 int n,t;
 double a1,a2;
 cin>>t;
 while(t--){
  cin>>n;
  a1=n*(log10(n*1.0));
  //cout<<a1<<endl;
  a2=a1-(__int64)a1;
  //cout<<a2<<endl;
  cout<<(int)pow(10.0,a2)<<endl;
 }
 return 0;
}

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