传送门：http://acm.hdu.edu.cn/showproblem.php?pid=5178

pairs

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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<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 x[i](?109≤x[i]≤109),
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

Source

BestCoder Round #31

题目给你N个点。问有多少对点的长度小于K

先快排。然后二分。。开始的时候没想到二分。。于是TLE了。。

后面没用lld 又WA了几次=。=这题真悲催。。

代码：

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<set>
#include<map>
#include<vector>
#include<cstring>
#include<stack>
#include<cmath>
#include<queue>
#define INF 0x0f0f0f0f
using namespace std;

int main()
{
	long long int i,j,k,l,r,n,t,m,mid;
	long long int a[100005];
	scanf("%d",&t);
	while(t--)
	{
		long long int ans=0;
		scanf("%I64d%I64d",&n,&k);
		for(i=0;i<n;i++)
		{
			scanf("%I64d",&a[i]);
		}
		sort(a,a+n);
		r=1;
		for(i=0;i<n;i++)
		{
			l=i+1;
			r=n-1;
			while(l<=r)
			{
				mid=(l+r)/2;
				if(a[mid]-a[i]>k)	r=mid-1;
				else l=mid+1;
			}

			ans=ans+r-i;

		}
		printf("%I64d\n",ans);
	}
}