HDU-1060 Leftmost Digit (数学)

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19262    Accepted Submission(s): 7632

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.

解答：

#include<iostream>  
#include<queue>
#include<cstdio>  
#include<cstring>  
#include<algorithm> 
#include<vector> 
#include<cmath>  
#include<sstream>
#include<cstdlib>
#include<map>
const int N =1005;
using namespace std;
int main(){
	int n,t;
	long long x;
	scanf("%d",&t);
	for(int i =0;i<t;i++){
		scanf("%d",&n);
		double m = n*log10((double)n);
		double r = m - (long long)m;//取小数部分
		r = pow(10.0,r);
		cout<<(int)r<<endl; 
	}
	return 0;
}