題鏈：http://codeforces.com/problemset/problem/251/A

A. Points on Line time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

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.

Sample test(s) 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}.

題意：找出三個，距離小於等於d的點，計算符合條件的三個點的所有組合數。

做法：從開始的點，固定一個點，然後二分找後面最大的但是二者距離不超過d的點，計算這些點c（n，2）。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <malloc.h>
#include <ctype.h>
#include <math.h>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#include <stack>
#include <queue>
#include <vector>
#include <deque>
#include <set>
#include <map>
#define INF 999999999
#define eps 0.00001
#define LL __int64
#define pi acos(-1.0)

int wei[100100];
int main()
{

	int n,d;
	int pre,id;
	__int64 num;
	__int64 ans;
	while(scanf("%d%d",&n,&d)!=EOF)
	{
		for(int i=0;i<n;i++)
		{
			scanf("%d",wei+i);
		}
		ans=0;
		for(int i=0;i<n;i++)
		{
			id=lower_bound(wei,wei+n,wei[i]+d)-wei;
			if(id==n)
				id--;
			else if(wei[id]!=wei[i]+d)
				id--;

			num=id-i;
			if(num>=2)
			{
				ans+=num*(num-1)/2; 
			}  
		} 
		printf("%I64d\n",ans);
	}
	return 0;
}
/*
100000 99

*/