HDU2578 Dating with girls(1)（二分）

没有灵魂的程序员

于 2017-07-17 12:28:58 发布

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分类专栏：二分查找文章标签： HDU

本文链接：https://blog.youkuaiyun.com/xky140610205/article/details/75223422

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本文介绍了一个经典的编程竞赛问题——寻找特定整数组合的解决方案数量。该问题要求在给定的一系列正整数中找到所有可能的配对，使得每一对整数之和等于指定的目标值。文章提供了一种高效的解决策略，利用排序和二分查找来减少计算时间。

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Dating with girls(1)

Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4870 Accepted Submission(s): 1557

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

HDU 2009-5 Programming Contest

http://acm.split.hdu.edu.cn/showproblem.php?pid=2578

题意很给力典型的二分注意位置不同也算一种相同数字去掉

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
int aa[100010];
int main()
{
    int t,n,k,i;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d",&n,&k);
        for(i=0; i<n; i++)
            scanf("%d",&aa[i]);
        sort(aa,aa+n);
        int  len=unique(aa,aa+n)-aa;
        long long sum=0;
        int l,r;
        for(i=0; i<len; i++)
        {
            l=0,r=len;
            int hj=k-aa[i];
            while(l<r)
            {
                int mid=(l+r)>>1;
                if(aa[mid]<hj) l=mid+1;
                else if(aa[mid]>hj)r=mid-1;
                else {l=mid; break;}
            }
            if(aa[l]==hj) sum++;
        }
        printf("%I64d\n",sum);
    }
    return 0;
}