hdu 5178 pairs (线性探查问题)

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

 题意：给定一个数组，求有多少对<a,b>使得|x[b]−x[a]|≤k.(a<b)

思路：1、这道题最重要的就是要解决超时这个问题，否则按暴力肯定是超时的。

        2、解决超时，这里有种方法。

            先将这个数组从小到大排序，然后i从0开始，tmp从0开始判断，如果值小于等于k的话，tmp++，直到>k停止，

            ans加上该值。下一次，i+1，之前的i和k可以，那么和i+1更可以，所有k继续向后跑，如此复杂度大大减少。

           具体看代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<cmath>
using namespace std;
#define N 100006
#define ll long long 
int n,k;
int a[N];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&k);
        for(int i=0;i<n;i++) scanf("%d",&a[i]);
        sort(a,a+n);
        ll ans=0;
        int tmp=0;
        for(int i=0;i<n;i++)
        {
            while(abs(a[i]-a[tmp])<=k && tmp<n ) tmp++;
            ans=ans+tmp-i-1;
        }
        printf("%I64d
",ans);
    }
    return 0;
}