cf Div#317 Lengthening Sticks-优快云博客

本文深入探讨了算法设计的核心原则与实践方法，通过详细分析排序算法、动态规划、哈希算法等关键技术，旨在帮助读者掌握高效解决问题的策略。同时，文章结合实例展示了贪心算法、深度优先搜索等经典算法的应用场景，为编程实践提供了宝贵的指导。

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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3跟长度为a、b、c的木条和长度为L的可增加量。可以给任意一根木条增加长度或不加，只要3个增加量的和不超过L，求增加后能使三根木条组成三角形的方法数。

下面为一位朋友的解题思路和代码，思路清晰而且代码简洁。

这道题差不多可以用容斥原理来做。

枚举长度（0-l）：

对于每一个长度，如果没有特殊情况，即(i+2)*(i+1)/2个。

但是若（其中一个+L）>（(a+b+c+i-1)/2）个，则减去(i-tmp+a+1)*(i-tmp+a)/2，另两个也是一样的。

至于为什么是(i+2)*(i+1)/2个，

如果我们想加上的总长度为i，则：
我们可以对a,b,c分别加上：
0,0,i 0,1,i-1, 0,2,i-2 0,3,i-3 ......
1,0,i-1 1,1,i-2 ......
i,0,0
是个三角状的。
======
=====
====
===
==
=
(形状差不多就是这样的）
所以是(i+2)*(i+1)/2。

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <queue>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <string>
#include <cstring>
using namespace std;

long long a,b,c,l;
long long Ans;

int main()
{
	scanf("%I64d%I64d%I64d%I64d",&a,&b,&c,&l);
	for(long long i=0;i<=l;i++)
	{
		long long tmp=(a+b+c+i-1)/2;//根据三角形的性质a<(c+b)
		if(a<=tmp&&b<=tmp&&c<=tmp)
		{
			Ans+=(i+2)*(i+1)/2;
			if(a+i>tmp)
			{
				Ans-=(i-tmp+a+1)*(i-tmp+a)/2;//减去不不能构成三角形的情况
			}
			if(b+i>tmp)
			{
				Ans-=(i-tmp+b+1)*(i-tmp+b)/2;
			}
			if(c+i>tmp)
			{
				Ans-=(i-tmp+c+1)*(i-tmp+c)/2;
			}
		}
	}
	
	printf("%I64d\n",Ans);
	return 0;
}