Question one hundred and nine: positive integer solution (2)

[Plain] Description
 
 
X + y + z = n is given a positive integer. n is used to calculate the number of positive integer solutions that meet the conditions.
 
 
Input
 
 
Each group of input data has a positive integer n (n> = 3)
 
 
Output
 
 
Output result
 
 
Sample Input
 
 
3
4
5
 
Sample Output
 
 
1
3
6

Description

X + y + z = n is given a positive integer. n is used to calculate the number of positive integer solutions that meet the conditions.

Input

Each group of input data has a positive integer n (n> = 3)

Output

Output result

Sample Input

3
4
5

Sample Output

1
3
6
 

[Plain] # include <stdio. h>
Int main ()
{
Int m;
Int n;
Int;
Int B;
Int c;

While (scanf ("% d", & n )! = EOF & n> = 3)
{
M = 0;
For (a = 1; a <n; a ++)
{
For (B = 1; B <n-a; B ++)
{
C = n-a-B;
If (c> = 0)
{
M ++;
}
}
}
 
Printf ("% d \ n", m );
}
 
Return 0;
}

# Include <stdio. h>
Int main ()
{
Int m;
Int n;
Int;
Int B;
Int c;

While (scanf ("% d", & n )! = EOF & n> = 3)
{
M = 0;
For (a = 1; a <n; a ++)
{
For (B = 1; B <n-a; B ++)
{
C = n-a-B;
If (c> = 0)
{
M ++;
}
}
}

Printf ("% d \ n", m );
}

Return 0;
}