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],),(i=,,,…,n−). 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 ), indicating the number of cases.
Each test case begins with two integers n,k(≤n≤,≤k≤).
Next n lines contain an integer x[i](−≤x[i]≤), 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

-     
5 300 
-    

 

Sample Output

 

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=;i<n;i++) scanf("%d",&a[i]);
         sort(a,a+n);
         ll ans=;
         int tmp=;
         for(int i=;i<n;i++)
         {
             while(abs(a[i]-a[tmp])<=k && tmp<n ) tmp++;
             ans=ans+tmp-i-;
         }
         printf("%I64d\n",ans);
     }
     return ;
 }