hdu--1060--Leftmost Digit

Leftmost Digit

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

Total Submission(s): 17741 Accepted Submission(s): 6854

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.

需要用到科学记数法和对数运算的知识。

我们把num*num的值记作：num*num=a*10^n，其中1

附上ac代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>

using namespace std;
typedef long long ll;

int main()
{
int t;

scanf ("%d",&t);

while (t--)
{
ll n;
int i,j;
double x;

scanf ("%lld",&n);
x = n*log10(n*(1.0));
x -= (ll)x;

int a = pow(10.0,x);

printf ("%d\n",a);

}
return 0;
}