pairs

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4207 Accepted Submission(s): 1504

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

Sample Output

3
10

思路：

把数组降序排序，这样我们从数组的第一个开始往后找就可以去掉绝对值符号了，对于每个xb,temp = xb - k < xa,只要二分找到恰好小于xa的那个temp,那么temp之前一直到xa的数就一定符合条件；

代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<map>
#include<set>
#include<cmath>
#define ll long long
#define pi acos(-1)
using namespace std;
int num[100005];
bool cmp(int x,int y)
{
   return x > y;
}
int main()
{
   int t;
   scanf("%d",&t);
   while (t --)
   {
       ll ans = 0;
       int n,k;
       scanf("%d %d",&n,&k);
       for (int i = 0;i < n;i ++)
           scanf("%d",&num[i]);
       sort(num,num + n,cmp);
       for (int i = 0;i < n - 1;i ++)
       {
           int temp = num[i] - k,mid;
           int l = i + 1,r = n - 1;
           while (l <= r)
           {
               mid = (l + r) / 2;
               if (num[mid] >= temp)
                   l = mid + 1;
               else
                   r = mid - 1;
           }
           ans += l - i - 1;
       }
       printf("%lld\n",ans);
   }
   return 0;
}

hdu5178 pairs 二分

pairs

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