HDU-1060-Leftmost Digit

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.

---

求n^n的第一位数。
令m=n^n，两边同取对数，得到log10(m)=n*log10(n)
再得到，m=10^(n*log10(n))
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
代码：

#include <iostream>#include <iomanip>#include <cstdio>#include <cstdlib>#include <string>#include <cstring>#include <cmath>#include <algorithm>#define N 1000010using namespace std;int main(){#ifndef ONLINE_JUDGE    freopen("1.txt", "r", stdin);#endif    int n, T;    long long t;    double a, b, c;    cin >> T;    while(T--)    {        cin >> n;        a = n*log10(n);        t = (long long)a;        b = a - t;        cout <<(long long)pow(10, b) << endl;    }    return 0;}