hdoj Leftmost Digit 1060 （數學取對數）

Leftmost Digit

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

Total Submission(s): 15786    Accepted Submission(s): 6154

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.
   
   
           
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<algorithm>
#define INF 0x3f3f3f3f
#define IN __int64
#define ull unsigned long long
#define ll long long
#define N 1010
#define M 1000000007
using namespace std;
int main()
{
	ll n;
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%lld",&n);
		double x=n*log10(n);
		x-=(ll)x;
		int y=pow(10.0,x);
		printf("%d\n",y);
	}
	return 0;
}