Lengthening Sticks - Codeforces 571 A

Lengthening Sticks

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given three sticks with positive integer lengths of a, b, and c centimeters. You can increase length of some of them by some positive integer number of centimeters (different sticks can be increased by a different length), but in total by at most l centimeters. In particular, it is allowed not to increase the length of any stick.

Determine the number of ways to increase the lengths of some sticks so that you can form from them a non-degenerate (that is, having a positive area) triangle. Two ways are considered different, if the length of some stick is increased by different number of centimeters in them.

Input

The single line contains 4 integers a, b, c, l (1 ≤ a, b, c ≤ 3·105, 0 ≤ l ≤ 3·105).

Output

Print a single integer — the number of ways to increase the sizes of the sticks by the total of at most l centimeters, so that you can make a non-degenerate triangle from it.

Sample test(s)

input

1 1 1 2

output

4

input

1 2 3 1

output

2

input

10 2 1 7

output

0

Note

In the first sample test you can either not increase any stick or increase any two sticks by 1 centimeter.

In the second sample test you can increase either the first or the second stick by one centimeter. Note that the triangle made from the initial sticks is degenerate and thus, doesn't meet the conditions.

题意：给你三个边的长度，和它们最多增加的长度，问有多少种方式使得其成为三角形。

思路：反过来求有多少种形式使得不是三角形。枚举三个边分别+i的长度且为最长边时的非法情况数目。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
int T,t;
ll solve(ll a,ll b,ll c,ll m)
{
    ll ret=min(a-b-c,m);
    if(ret<0)
      return 0;
    else
      return (ret+2)*(ret+1)/2;
}
int main()
{
    int i,j,k;
    ll a,b,c,n,ans;
    scanf("%I64d%I64d%I64d%I64d",&a,&b,&c,&n);
    ans=(n+3)*(n+2)*(n+1)/6;
    for(i=0;i<=n;i++)
    {
        ans-=solve(a+i,b,c,n-i);
        ans-=solve(b+i,a,c,n-i);
        ans-=solve(c+i,a,b,n-i);
    }
    printf("%I64d\n",ans);
}