HDU 1060 Leftmost Digit

Problem Description

Given a positive integer N, you should output the leftmost digit of N^N.



[b]Input
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[b]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).
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[b]Output
[/b]

For
each test case, you should output the leftmost digit of N^N.




[b]Sample
Input
[/b]

2

3

4




[b]Sample
Output
[/b]

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.

总结1：题意理解：输出 N的N次方 的 最左边 的 第一个数字。

总结2：r=（int)pow(10,n*log10(n)-(int)(n*log10(n)));

总结3：公式推导:
设n^n=a0*10^m+a1*10^(m-1)+...a0,a1...为相应位的系数,

m为数字位个数,

如4^4=256,a0=2,a1=5,a2=6,m=3；　　

　a0*10^m<=n^n<(a+1)*10^m两边取对数m+lga0<=nlgn<m+lg(a0+1)

即lga0<=nlgn - m<lg(a0+1)

所以a0<=10^(nlgn-m)<a0+1；

　　1<=a0<=9；

所以0<=lga0<1；

由m+lga0<=nlgn<m+lg(a0+1)两边取整得m=[nlgn](表示nlgn的整数部分)

所以a0=[10^(nlgn-[nlgn])]=pow(10,n*log10(n)-(int)n*log10(n))