Leftmost Digit

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

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.


 

Author

Ignatius.L

一看题上的 数值范围 就知道一般的数值类型没法表示 考虑到__int64

又因为n^n数值太大 考虑用指数表示法

则有n^n=a*10^x;

等号两边对10取对数得 n*log10n=lga+x;

则lga=(n*lgn-x)

x可以通过对n*lgn向下取整求得

则有lga=(n*lgn-[n*lgn])

所以有以下代码：

#include<stdio.h>

#include<math.h>

int main()

{

int t;

scanf("%d",&t);

while(t--)

{

__int64 n,s;

double m;

scanf("%I64d",&n);

m=n*log10(n+0.0);

m-=(__int64)m;

s=(__int64)pow(10,m);

printf("%I64d\n",s);

}

return 0;

}

杭电 1060