Leftmost Digit（数学）

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. 

题目意思：求n^n结果的第一位。
解题思路：我们可以将其看成幂指函数，那么我们对其处理也就顺其自然了，n^n = 10 ^ (log10(n^n)) = 10 ^ (n * log10(n)),
然后我们可以观察到： n^n = 10 ^ (N + s) 其中，N 是一个整数 s 是一个小数。由于10的任何整数次幂首位一定为1,所以首位只和s(小数部分)有关。 

 #include<cmath>
 #include<cstdio>
 #include<algorithm>
 #define ll long long int
 using namespace std;
 int main()
 {
     int t,n;
     double m;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d",&n);
         m=n*log10(n);
         m=m-(ll)(m);
         m=pow(10.0,m);
         printf("%d\n",(int)(m));
     }
     return ;
 }