cf 251 A Points on Line 二分

题链：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 31 2 3 4

output

4

input

4 2-3 -2 -1 0

output

2

input

5 191 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：找出三个，距离小于等于d的点，计算符合条件的三个点的所有组合数。

做法：从开始的点，固定一个点，然后二分找后面最大的但是二者距离不超过d的点，计算这些点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*/