The number of hdu-2502-months

Topic links

http://acm.hdu.edu.cn/showproblem.php?pid=2502

Analysis:

For example, when n=4, there are:

1000 1001 1010 1011 1100 1101 1110 1111 you can see. Except for the first one. There are 000 001 010 011 100 101 110 111 can count, 0 and 1 of the total number half to half. So calculate the total number of 0 and 1, divided by 2 is good.

That is: 2^ (n-2) * (n-1)

Plus the first digit 1 total number: 2^ (n-1) The number of months

i.e. answer = 2^ (n-2) * (n-1) + 2^ (n-1)

Code

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>

int main (void)
{
int t,n;
scanf ("%d", &t);
while (t--)
{
scanf ("%d", &n);
printf ("%.lf\n", pow (2,n-2) * (n-1) +pow (2,n-1));
}
return 0;
}

The number of hdu-2502-months