n=int(input())
ans=0;
for i in range(n):
    for j in range(i,n):
        k=n-i-j
        if(k<j):#循环枚举，利用if语句使k>j>i防止重复计数
            continue
        if(i+j>k):
            ans+=1

print(ans)

l=int(input())
a=b=n=0
for a in range(1,l):
for b in range(1,l-a):
c=l-a-b
if b<=a and c<=b:
if a+b>c and a+c>b and b+c>a:
n=n+1
print(n)

一个比较奇怪的做法
n=int(input())
p=(n+1)//2
k=0
for a in range(0,p):
for b in range(0,p):
c=n-a-b
if a+b>c and c>0 and b>0 and a>0:

if a==b:
if b==c:
k=k+6
else:
k=k+2
elif a==c and c!=b:
k=k+2
elif c==b and c!=a:
k=k+2
else:
k=k+1
p=k//6
print(p)

#include <iostream>
#include <stdlib.h>

using namespace std;

int main()
{
    int n;
    cin >> n;
    int ans = 0;
    for(int i = 1; i <= n/2 - (!(n % 2)); i++)
    {
        for(int j = 1; j <= i; j++)
        {
            int k = n - i - j;
            if(k > j)
            {
                continue;
            }
            else
            ans++;
        }
    }
    cout << ans;
    return 0;
}

n=int(input())
cnt=0
for i in range(n//2+1):
    for j in range(i,n//2+1):
        if i+j>n-i-j and n-i-j>=j:
            cnt+=1
print(cnt)

#include <stdio.h>
#include<iostream>
using namespace std;
int n,cnt;
int main()
{
    cin>>n;
    for(int i=1;i<=n/2-(!(n%2));i++)
    {
        for(int j=1;j<=i;j++)
        {
            int k=n-i-j;
            if(k>j)continue;
            cnt++;
        } 
    }   
    cout<<cnt; 
}