Hdu 1060 Leftmost Digit

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18251    Accepted Submission(s): 7146

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.

#include<stdio.h>
#include<math.h>int main()
{
    __int64 cas,b,i,d;
    double a,m,n,c;
    scanf("%I64d",&cas);
    for(i=1;i<=cas;i++)
    {
        scanf("%lf",&n);
        a=n*log10(n);
        b=(__int64)(a);
        c=a-b;
        d=(__int64)(pow(10,c));
        printf("%I64d
",d);
    }
    return 0;
}