 # Problem Triangle creation.What's wrong with my logic.Help

What is wrong with my logic and code.Help please.Thanks.

You are given a stick of length N. You want to break it in three pieces such that it can form a triangle. How many distinct triangles can you make? Two triangles are equal if all the side lengths are same when sorted in ascending order of length. So (1, 3, 2) is same to (3, 1, 2) because their side lengths are same if we sort them, which is (1, 2, 3). But (1, 3, 4) is not same with (1, 2, 3). Suppose the lengths of three pieces are X, Y, Z (X <= Y <= Z) respectively. Following constraints should be maintained:

1. X, Y, Z > 0.
2. X, Y, Z is an integer.
3. X + Y >= Z
4. X + Y + Z = N
For example if N = 14, then there are 7 triangles: (1, 6, 7), (2, 5, 7), (2, 6, 6), (3, 4, 7), (3, 5, 6), (4, 4, 6), (4, 5, 5).
Input
First line will give you the number of test cases, T (T<=100). Then each line will have an integer N (0< N <= 300000).
Output
For each test case, print the test case number starting from 1 and an integer denoting the number of distinct triangles possible.
Sample Input
Output for Sample Input
3 3 6 14
Case 1: 1
Case 2: 2
Case 3: 7

#include<stdio.h>
#define max 40
void main()

{

int q=0,m=0,i=0,j=0,k=0,n=14,p,count=0,fcount=0,tri[max];
//printf(“Enter n:”);
//scanf("%d",&n);

for(j=1;j<(n-1);j++)

{for(k=1;k<=j;k++)

{

if((j+k)>(n-(j+k))&& n!=(j+k) && (((n-(j+k))+k) >j) && ((n-(j+k))+j)>k)

{p=n-(j+k);

if(p>=j && p>=k)

{tri[i]=p;tri[i]=j;tri[i]=k;}

else if(p<=j && p<=k)

{tri[i]=j;tri[i]=k;tri[i]=p;}
else if(p<=j && p>=k)

{tri[i]=j;tri[i]=p;tri[i]=k;}

i++;
count++;

}

}

}

fcount=count;

for(m=0;m<(count);m++)
for(q=(count-1);q>m && q!=m;q–){
{
if(tri[m]==tri[q])
{
if(tri[m]==tri[q])
{
if(tri[m]==tri[q])
{fcount–;}
}
}

}
}
printf("%d\n%d",count,fcount);
}