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

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<stdlib.h>
 6 #include<cmath>
 7 using namespace std;
 8 #define N 100006
 9 #define ll long long
10 int n,k;
11 int a[N];
12 int main()
13 {
14     int t;
15     scanf("%d",&t);
16     while(t--)
17     {
18         scanf("%d%d",&n,&k);
19         for(int i=0;i<n;i++) scanf("%d",&a[i]);
20         sort(a,a+n);
21         ll ans=0;
22         int tmp=0;
23         for(int i=0;i<n;i++)
24         {
25             while(abs(a[i]-a[tmp])<=k && tmp<n ) tmp++;
26             ans=ans+tmp-i-1;
27         }
28         printf("%I64d\n",ans);
29     }
30     return 0;
31 }