hdu 5178 pairs（尺取法求解）

pairs

Problem Description

John has n points on the X axis, and their coordinates are (x[i],0),(i=0,1,2,…,n−1). He wants to know how many pairs 《a,b》 that |x[b]−x[a]|≤k.(a小于b)

Input

The first line contains a single integer T (about 5), indicating the number of cases.

Each test case begins with two integers n,k(1≤n≤100000,1≤k≤109).

Next n lines contain an integer xi, means the X coordinates.

Output

For each case, output an integer means how many pairs《a,b》 that |x[b]−x[a]|≤k.

Sample Input

2

5 5

-100

0

100

101

102

5 300

-100

0

100

101

102

Sample Output

3

10

解析：本题还除了用二分还可以用尺取法求解，注意超时，开一个下标变量从1一次性走到n，因为可以利用每次循环过的以前的条件，比如3~6满足条件，当i走到4的时候，j就不必要从5开始而是直接从7开始走了。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
long long a[100005];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
long long n,k,cnt=0,r,l,sum,mid=2;
scanf("%lld %lld",&n,&k);
for(int i=1;i<=n;i++)
scanf("%lld",&a[i]);
sort(a+1,a+n+1);
for(int i=1,j=1;i<=n;i++)        //变量j从1走到n，时间复杂度为O（n）
{
while(a[j+1]-a[i]<=k)   //循环条件
{
if(j+1>n)     //边界判断
break;
j++;
}
cnt+=j-i;       //计数变量逐次递加
}
printf("%lld\n",cnt);
}
return 0;
}