C - Count Triangles

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本文介绍了一种算法，用于计算给定整数范围内满足特定条件的非退化三角形的数量。该算法通过枚举三角形两边之和并计算第三边的有效范围来实现。

Like any unknown mathematician, Yuri has favourite numbers: A, B, C, and D, where A≤B≤C≤D. Yuri also likes triangles and once he thought: how many non-degenerate triangles with integer sides x, y, and z exist, such that A≤x≤B≤y≤C≤z≤D holds?

Yuri is preparing problems for a new contest now, so he is very busy. That’s why he asked you to calculate the number of triangles with described property.

The triangle is called non-degenerate if and only if its vertices are not collinear.

Input
The first line contains four integers: A, B, C and D (1≤A≤B≤C≤D≤5⋅105) — Yuri’s favourite numbers.

Output
Print the number of non-degenerate triangles with integer sides x, y, and z such that the inequality A≤x≤B≤y≤C≤z≤D holds.

Examples
Input
1 2 3 4
Output
4
Input
1 2 2 5
Output
3
Input
500000 500000 500000 500000
Output
1
Note
In the first example Yuri can make up triangles with sides (1,3,3), (2,2,3), (2,3,3) and (2,3,4).

In the second example Yuri can make up triangles with sides (1,2,2), (2,2,2) and (2,2,3).

In the third example Yuri can make up only one equilateral triangle with sides equal to 5⋅105.

首先固定x+y为i那么z的取值范围便是c~min(d,i-1)；
而对于x和y的范围，当i<=2*b时固定y为b则x=i-b（大于同理），然后减小x的值在固定i的条件下y就要增加，这样x和y的最小活动范围长度就是x和y的方案数
令x和y的方案数与z的方案数相乘便得到了x+y==i的方案数
枚举i的取值区间并加和即可得到所有方案数

#include<iostream>
using namespace std;
int main()
{
	int a,b,c,d,x,y,z;
	long long ans=0;
	cin>>a>>b>>c>>d;
	for(int i=a+b;i<=b+c;i++)
	{
		z=min(d,i-1);
		if(z<c) continue;
		z=z-c+1;
		if(i<=2*b)
		{
			x=i-b;
			ans+=1ll*(min(x-a,c-b)+1)*z;
		}
		else
		{
			y=i-b;
			ans+=1ll*(min(b-a,c-y)+1)*z;
		}
	}
	cout<<ans<<endl;
}