Dating with girls(1)

最新推荐文章于 2021-08-18 19:29:53 发布

userluoxuan

最新推荐文章于 2021-08-18 19:29:53 发布

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分类专栏：解题报告二分 ====搜索==== 文章标签：二分

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此博客讨论了一个关于在给定整数集合中寻找满足特定条件配对的问题，即找到一对数字使得它们之和等于给定的目标值，并且考虑到了整数的顺序差异和重复值的处理。

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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

解题思路：

题目大意是给定一组数，让你选定两个数，使之相加等于k。选择的数字可以重复使用，选的数字顺序不同记为两种情况。让你求出所有的情况。先对这组数进行排序，然后遍历一遍进行折半查找即可。题目易出错的地方在于给的数字可能是重复的，需要去重处理。我其实把代码编的复杂了，完全可以查找的时候如果这组数出现的数字重复就跳过该数。而我用了vector进行去重处理，真是no zuo no die 啊。。。。。给的代码有多处还可以进行优化。。。。

AC代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
const int maxn = 100005;
int main()
{
    int t, n, k;

    scanf("%d", &t);
    while(t--)
    {
        vector<int> a;
        __int64 ans = 0;
        int b;
        bool flag = false;
        scanf("%d%d", &n, &k);
        for(int i = 0; i < n; i++)
        {
            scanf("%d", &b);
            a.push_back(b);
            if(2 * b == k)
                flag = true;
        }
        sort(a.begin(), a.end());
        vector<int> :: iterator it = unique(a.begin(), a.end());  // 傻乎乎的进行了去重处理。。。。
        a.erase(it, a.end());
        for(int i = 0; i < a.end() - a.begin(); i++)
        {
            int low = i, high = a.end() - a.begin() - 1, mid;
            if(a[i] >= k)
                break;
            while(high >= low)
            {
                mid = (high + low) / 2;
                if(a[i] + a[mid] > k)
                    high = mid - 1;
                if(a[i] + a[mid] < k)
                    low = mid + 1;
                if(a[i] + a[mid] == k)
                {
                    ans++;
                    break;
                }
            }
        }
        if(flag)
            printf("%I64d\n", ans * 2 - 1);
        else
            printf("%I64d\n", ans * 2);
    }
    return 0;
}