* The leftmost number

Problem S
Time limit:1000 MS Memory limit:32 MB 64bit IO Format:%i64d
submitted:61 accepted:25
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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 was a single integer T which is the number of test cases. T test Cases follow.
Each test case is 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 = leftmost, so the-the-digit is 2.
In the second case, 4 * 4 * 4 * 4 = at the very leftmost digit is 2.

 [CPP] View plain copy on code to view the snippet derivation to my code slice 01./* the idea: 02.1, make m=n^n; 03. 
04.2, the log10 (M) =n*log10 (n), the m=10^ (N*LOG10 (n)), and 05 are taken on both sides of the equivalence. 
06.3, so that n*log10 (N) =a+b,a is an integer, B is a decimal, and 07. 
08.4,c function: log10 (), calculates logarithm, pow (A, b) calculates a^b 09. 
10.5, because 10 of any integer power of the first must be 1, so, the first of M and N*log10 (N) of the small part of the section, 11.    
12. That is, only by seeking 10^b to save; 13. 14.6, the final 10^b rounding, the output of the number of rounding is OK. 
(Because 0<=b<1, so 1<=10^b<10 to 15. 
16. Its rounding, then is a bit, that is, the number of the request).   

17.*/18.//hdoj system is a pit, all kinds of CE, all kinds of WA, thanks to Kangxiaohui guidance.
    #include <stdio.h> #include <math.h> int main () {long n,i,s,m,a;
    scanf ("%i64d", &m);
     for (i=0;i<m;i++) {scanf ("%i64d", &n);
     a= (int) (POW (10,N*LOG10 (n)-(Long Long) (N*LOG10 (n)));
    printf ("%i64d\n", a);
} return 0; }