C. Count Triangles
链接：https://codeforces.com/contest/1355/problem/C
Like any unknown mathematician, Yuri has favourite numbers: AA, BB, CC, and DD, where A≤B≤C≤DA≤B≤C≤D. Yuri also likes triangles and once he thought: how many non-degenerate triangles with integer sides xx, yy, and zz exist, such that A≤x≤B≤y≤C≤z≤DA≤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: AA, BB, CC and DD (1≤A≤B≤C≤D≤5⋅1051≤A≤B≤C≤D≤5⋅105) — Yuri's favourite numbers.
Output
Print the number of non-degenerate triangles with integer sides xx, yy, and zz such that the inequality A≤x≤B≤y≤C≤z≤DA≤x≤B≤y≤C≤z≤D holds.
Examples
input
Copy
1 2 3 4
output
Copy
4
input
Copy
1 2 2 5
output
Copy
3
input
Copy
500000 500000 500000 500000
output
Copy
1
Note
In the first example Yuri can make up triangles with sides (1,3,3)(1,3,3), (2,2,3)(2,2,3), (2,3,3)(2,3,3) and (2,3,4)(2,3,4).
In the second example Yuri can make up triangles with sides (1,2,2)(1,2,2), (2,2,2)(2,2,2) and (2,2,3)(2,2,3).
In the third example Yuri can make up only one equilateral triangle with sides equal to 5⋅1055⋅105.
qlsNB！！！
代码：
#include <algorithm>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <clocale>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <limits>
#include <list>
#include <map>
#include <iomanip>
#include <ios>
#include<iosfwd>
#include <iostream>
#include <istream>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <utility>
#include <vector>
#include <cwchar>
#include <cwctype>
#define ll long long
#define rep(i,a,n) for(int i=a;i<=n;i++)
using namespace std;
ll n,m,t,k,s,z,y,l,r,ans;
ll a,b,c,d;
int main()
{
cin>>a>>b>>c>>d;
ans=0;
for(ll i=max(a+b,c+1);i<=b+c;i++)
{
l=max(a,i-c);
r=min(b,i-b);
if(l<=r)
{
ans+=(r-l+1)*(min(i-1,d)-c+1);
}
}
cout<<ans;
}
