hdu 2578——Dating with girls(1)

本文详细介绍了如何解决HDU2009-5编程竞赛中的一道问题,即求解给定整数集合中满足特定条件的方程组的解的数量。通过输入一组整数和目标值,读者将学习到一种高效的方法来计算符合条件的解的数量,并通过实例展示了实际应用。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

原题

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
 


 

Source

 

源码:

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int a[100005];
int main()
{
    int t;
    int n,m;
    int i,j;
    int k;
    cin>>t;
    while(t--)
    {
        cin>>n>>k;
        for(i=1; i<=n; i++)
            scanf("%d",&a[i]);
        a[0]=-99999;
        sort(a,a+n+1);
        int count=0;
        int big,small;
        int mid;
        for(i=1; i<=n; i++)
        {
            if(a[i-1]!=a[i])
            {
                big=n;
                small=1;
                while(small<=big)
                {
                    mid=(big+small)/2;
                    if(a[mid]+a[i]==k)
                    {
                        count++;
                        break;
                    }
                    if(a[mid]+a[i]>k)
                        big=mid-1;
                    else
                        small=mid+1;
                }
            }
        }
        cout<<count<<endl;
    }
    return 0;
}


 

 
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值