HDU 1060 leftmost digit__ winter training

Problem Description Given A positive integer N, you should output the leftmost digit of n^n.
Input the input contains several test cases. The ' the ' 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 "the", 3 * 3 * 3 = leftmost digit is 2. In the second case, 4 * 4 * 4 * 4 = 256, so the leftmost digit is 2. M=n^n, the two sides respectively to the 10 logarithm of the log10 (m) =n*log10 (n), m=10^ (N*LOG10 (n)), since the first of the 10 of any integer power must be 1, so the first of M and N*log10 (n) of the number of parts.

#include <cstdio>
#include <math.h>
int main ()
{
    int t;
	scanf ("%d", &t);
        while (t--)
        {
            unsigned long n;
            scanf ("%lld", &n);
            Double X=n*log10 (n*1.0);
            x-= (__int64) x;
            int A=pow (10.0, x);
            printf ("%d\n", a);
        }
    return 0;
}