hdu 5178 pairs

pairs

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

Total Submission(s): 320 Accepted Submission(s): 143



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

真的是很久很久没有用到二分查找了，真的是没想到，想了半天都没想到这个。。。。

这里我们需要注意的一点是，数据比较大，要用long long，不能用int。

我们遍历每一个a[i].然后找到位子极限位子r，使得r-i个a都能满足这个条件，然后依次相加得出答案。这里我们直接上代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define ll long long int
ll a[100005];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        ll n,k;
        scanf("%I64d%I64d",&n,&k);
        for(int i=0;i<n;i++)
        {
            scanf("%I64d",&a[i]);
        }
        sort(a,a+n);
        ll output=0;
        for(int i=0;i<n;i++)//依次遍历每一个a【i】，找到他们的极限位子r
        {
            int l=i+1;
            int r=n-1;
            int mid=(l+r)/2;
            while(l<=r)
            {
                mid=(l+r)/2;
                if(a[mid]-a[i]>k)//如果中间位子的a【mid】-a【i】>k明显我们要找的位子要在mid左边，所以我们这里让r=mid-1就行了
                r=mid-1;
                else
                l=mid+1;//这里同理
            }
            output+=r-i;
            //printf("%d\n",r);
        }
        printf("%I64d\n",output);
    }
}