Codeforces 251A+（五一训练 C）+二分法

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
%I64d specifier.

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}.

分析；n个点，距离d，找出集合中每三点最远和远近点距离不超过d的集合。

首先确定起点，然后二分确定最大点，从而确定了一个范围，第二小点就在最大点与最小点间选择。

这题和山大的那道三角形的方法是一样的。

#include<iostream>
#include<string.h>
#include<stdio.h>
#include<algorithm>
using namespace std;
long long num[100005],n,d;
int main()
{
    long long i,ans=0;
    cin>>n>>d;
    for (i = 0; i < n; i ++)
        cin>>num[i];
    for (i = 0; i < n - 2 ; i ++)
    {
    long long mid,start=i,a=i+2,b=n-1;
    while (a < b)
    {
        mid = (a + b) / 2;
        if (num[mid]  - num[start] <= d)
            a = mid + 1;
        else
            b = mid;
    }
    mid = (a + b) / 2;
    if (num[mid] - num[start] > d)
        mid --;
        if (mid - i < 2) continue;
        ans += (mid- i) * (mid- i - 1) / 2;
    }
    cout<<ans<<endl;
    return 0;
}