HDOJ 2578 Dating with girls(1)

Dating with girls(1)

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4243    Accepted Submission(s): 1326

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
   
   
   
   

 

   

  此题主要考察了二分查找的运用，经过此题我发现自己还很嫩啊，要好好学扎实了才行！

  注意相同的数字只算一个，要排除相同的

  代码如下

#include<cstdio>
#include<algorithm>
using namespace std;
int Binary_Search(int a[],int n,int k)
{
    int low,high,mid,counter=0;
    for(int i=0;i<n;i++)
    {
        if(a[i]==a[i+1])
            continue;
        low=0,high=n-1;
         while(low<=high)
        {
            mid=(low+high)/2;
            if(a[mid]+a[i] > k)
                high = mid -1;
            else if(a[mid]+a[i] < k)
                low = mid +1;
            else
            {
                counter++;
                break;
            }
        }
    }
    return counter;
}
int main()
{
    int i,n,t,k,counter;
    int a[100005];
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&k);
        for(i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
       sort(a,a+n);//用sort函数进行升序
       counter=Binary_Search(a,n,k);
       printf("%d\n",counter);
    }
    return 0;
}