C - Count Triangles

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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;
}
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CodeForces - 1355C Count Triangles(数学)

题目链接:点击查看 题目大意:给出 A B C D ,规定 A <= x <= B <= y <= C <= z <= D ,问 ( x , y , z ) 三元组为三角形的情况有多少种 题目分析:首先需要知道,当 x + y > z 时,三元组 ( x , y , z ) 满足条件,但是三个数的范围都是 1e5 级别的,不能直接暴力枚举 但是不难想到可以 O( n ) 枚举 x ,然后尝试 O( 1 ) 去计算 ( y , z ) 有多少对适应当前的 x

Falcon的博客 983

CF1355C Count Triangles

首先x+y一定要大于z才能构成,所以一定要大于c(因为c是z的最小取值),统计答案的时候呢,最普通的就是x+y-c,(假设x+y是5,c是3,那么z的取值是3,4,所以是x+y-c,不加1,否则z取5就是两边之和等于第三边了)。所以我们统计x+y一共出现了多少次,考虑差分,枚举x,也就是将x+c到x+d之间的数都加上1,也就是s[x+c]++,s[x+d+1]--.再做一遍前缀和复原,在统计x+y的答案时乘上s[x+y]就好了。注意判断的时候一定是x+y>c,如果写等于cf数据也是可以过的。

wenhaorjxj的博客 321

C. Count Triangles -差分 -枚举

考虑最为暴力的方法,也就是统计x+y>z的数量, 枚举z,对于每一个z,暴力枚举大于z的x+y,显然会超时。值得注意的是z1

jisuanji2606414的博客 242

Codeforces1355 C. Count Triangles(枚举)

题意: 给定A、B、C、D 给定限定条件A<=x<=B<=y<=C<=z<=D 问有多少个(x,y,z)能组成三角形 数据范围:A,B,C,D<=5e5 解法: 题目即统计满足x+y<z的方案数 容易想到枚举一个,O(1)计算方案数,但是细节有点多,不好码 还有一种做法是枚举x+y,设当前枚举到的x+y为i 那么需要计算的有两个: 1.满足i<z的z的数量,这个非常简单 2.满足x<=y且x+y=i的(x,y)的数量,这个稍微麻烦一点 考虑如何计算

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C. Count Triangles(数学分情况推式子)

传送门 思路清晰分好情况,就不难思路清晰分好情况,就不难思路清晰分好情况,就不难 三条边的区间分别是[a,b],[b,c],[c,d]三条边的区间分别是[a,b],[b,c],[c,d]三条边的区间分别是[a,b],[b,c],[c,d] 我们枚举第一条边[a,b],算出每种情况的答案。我们枚举第一条边[a,b],算出每种情况的答案。我们枚举第一条边[a,b],算出每种情况的答案。 一、枚举第一条边A后,第二条B有完全符合的\color{Red}一、枚举第一条边A后,第二条B有完全符合的一、枚举第一条边A后

jziwjxjd的博客 546
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