Codeforces 251A-Points on Line

最新推荐文章于 2022-01-20 16:06:58 发布

Wang_128

最新推荐文章于 2022-01-20 16:06:58 发布

阅读量774

点赞数

CC 4.0 BY-SA版权

分类专栏： Codeforces ----二分 ----其他数论

本文链接：https://blog.youkuaiyun.com/a664607530/article/details/56016281

Codeforces 同时被 3 个专栏收录

140 篇文章

订阅专栏

----二分

44 篇文章

订阅专栏

----其他数论

15 篇文章

订阅专栏

本文探讨了一个经典的组合数学问题，即从一组沿着Ox轴分布的点中选择三个点的方法数量，使得最远两点间的距离不超过给定值d。文章提供了完整的代码实现，包括输入输出示例，帮助读者理解算法细节。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Little Petya likes points a lot. Recently his mom has presented him n points lying on the line OX. Now Petya is wondering in how many ways he can choose three distinct points so that the distance between the two farthest of them doesn't exceed d.

Note that the order of the points inside the group of three chosen points doesn't matter.

Input

The first line contains two integers: n and d (1 ≤ n ≤ 105; 1 ≤ d ≤ 109). The next line contains n integers x1, x2, ..., xn, their absolute value doesn't exceed 109 — the x-coordinates of the points that Petya has got.

It is guaranteed that the coordinates of the points in the input strictly increase.

Output

Print a single integer — the number of groups of three points, where the distance between two farthest points doesn't exceed d.

Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64dspecifier.

Examples

input
4 3
1 2 3 4

output
4

input
4 2
-3 -2 -1 0

output
2

input
5 19
1 10 20 30 50

output
1

Note

In the first sample any group of three points meets our conditions.

In the seconds sample only 2 groups of three points meet our conditions: {-3, -2, -1} and {-2, -1, 0}.

In the third sample only one group does: {1, 10, 20}.

题意：问从集合中取出3个数，最大数与最小数的差值小于k的方式有多少种。

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits>

using namespace std;

#define LL long long
const int INF=0x3f3f3f3f;
const int MAXN=1000010;

int a[MAXN];

int main()
{
    int n,k;
    while(~scanf("%d %d",&n,&k))
    {
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        LL ans=0;
        for(int i=1;i<=n-2;i++)
        {
            LL p=upper_bound(a+i,a+1+n,a[i]+k)-a-1;
            ans+=(p-i)*(p-i-1)/2;
        }
        printf("%lld\n",ans);
    }
    return 0;
}