HDOJ-2578-Dating with girls(1) 【排序 去重 二分】

Dating with girls(1)

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4042    Accepted Submission(s): 1248

Problem Description

Everyone in the HDU knows that the number of boys is larger than the number of girls. But now, every boy wants to date with pretty girls. The girls like to date with the boys with higher IQ. In order to test the boys ' IQ, The girls make a problem, and the boys who can solve the problem
correctly and cost less time can date with them.
The problem is that : give you n positive integers and an integer k. You need to calculate how many different solutions the equation x + y = k has . x and y must be among the given n integers. Two solutions are different if x0 != x1 or y0 != y1.
Now smart Acmers, solving the problem as soon as possible. So you can dating with pretty girls. How wonderful!

 

Input

The first line contain an integer T. Then T cases followed. Each case begins with two integers n(2 <= n <= 100000) , k(0 <= k < 2^31). And then the next line contain n integers.

 

Output

For each cases,output the numbers of solutions to the equation.

 

Sample Input

  
  

   
   2
5 4
1 2 3 4 5
8 8
1 4 5 7 8 9 2 6
  
  

 

Sample Output

  
  

   
   3
5
  
  

恩，题目大意就是说，给 n 个数和 k ，然后从这 n 个数中找两个数相加和为 k ，问共有多少种组合，例如 3 + 2 = 5 和 2 + 3 = 5 为两种，但 4 + 4 = 8 就一种。开始直接 for 循环写，结果超时，看讨论区，要去重，还是会超时，所以要二分查找。

#include <iostream>
#include<cstdio>
#include<cstring>
#define maxn 100000+10
#include<algorithm>
using namespace std;
int rec[maxn];
int n;
int bsearch(int s)
{
    int l=0,r=n-1,mid;
    while(l<=r)
    {
        mid=(l+r)>>1;
        if(rec[mid]==s)
            return 1;
        if(s<rec[mid])
            r=mid-1;
        else
            l=mid+1;
    }
    return 0;
}
int main()
{
    int t,k,cnt,flag;
    scanf("%d",&t);
    while(t--)
    {
        flag=cnt=0;
        scanf("%d%d",&n,&k);
        for(int i=0;i<n;++i)
            scanf("%d",&rec[i]);
        sort(rec,rec+n);
        rec[n]=-1;
        for(int i=0;i<n;++i)
        {
            if(rec[i]!=rec[i+1]&&bsearch(k-rec[i]))
            {
                cnt++;
            }
        }
        printf("%d\n",cnt);
    }
    return 0;
}