Leftmost Digit

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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

Sample Output

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.

题目大意：算出N^N最左边的数字是多少，直接算肯定溢出，

m=n^n;两边同取对数，得到，log10(m)=n*log10(n);再得到，m=10^(n*log10(n));

然后，对于10的整数次幂，第一位是1，所以，第一位数取决于n*log10(n)的小数部分（涉及了对数运算，因为整数部分的为10^m,对最左边位置数无影响）

#include <iostream>#include<cstdio>#include<cmath>using namespace std;int main(){    int icase;    scanf("%d",&icase);    while(icase--)    {        int n;        scanf("%d",&n);        double r1=n*log10(n);        double r2=r1-(long long)r1;        printf("%d\n",(int)pow(10,r2));    }}

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