HDU5857-Median

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本文解析了一道算法题目，该题目要求在给定的有序序列中通过多次查询找到两个子序列合并后的中位数。文章提供了详细的解题思路及实现代码，适合对算法竞赛感兴趣或希望提高算法能力的读者。

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Median

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1563 Accepted Submission(s): 417

Problem Description

There is a sorted sequence A of length n. Give you m queries, each one contains four integers, l1, r1, l2, r2. You should use the elements A[l1], A[l1+1] ... A[r1-1], A[r1] and A[l2], A[l2+1] ... A[r2-1], A[r2] to form a new sequence, and you need to find the median of the new sequence.

Input

First line contains a integer T, means the number of test cases. Each case begin with two integers n, m, means the length of the sequence and the number of queries. Each query contains two lines, first two integers l1, r1, next line two integers l2, r2, l1<=r1 and l2<=r2.
T is about 200.
For 90% of the data, n, m <= 100
For 10% of the data, n, m <= 100000
A[i] fits signed 32-bits int.

Output

For each query, output one line, the median of the query sequence, the answer should be accurate to one decimal point.

Sample Input

Sample Output

  
   2.0
1.5

Author

BUPT

Source

2016 Multi-University Training Contest 10

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wange2014

题意：给你一个有序序列，有q次询问，每次给出l1,r1和l2,r2，序列l1~r1和序列l2~r2形成一个新序列，求出新序列的中间值

解题思路：暴力判断

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <map>
#include <cmath>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int n,m,l1,l2,r1,r2;
LL a[100009];

void get(int p,int &x)
{
    if(l2>r1)
    {
        if(p+l1-1<=r1) x=p+l1-1;
        else x=p-(r1-l1+1)+l2-1;
    }
    else if(r1<=r2)
    {
        if(p<=l2-l1) x=p+l1-1;
        else if(p>r1-l1+1+r1-l2+1) x=p-(r1-l1+1+r1-l2+1)+r1;
        else x=(p-(l2-l1)+1)/2+l2-1;
    }
    else
    {
        if(p<=l2-l1) x=p+l1-1;
        else if(p>r2-l1+1+r2-l2+1) x=p-(r2-l1+1+r2-l2+1)+r2;
        else x=(p-(l2-l1)+1)/2+l2-1;
    }
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(int i=1; i<=n; i++) scanf("%lld",&a[i]);
        while(m--)
        {
            scanf("%d%d%d%d",&l1,&r1,&l2,&r2);
            if(l1>l2)
            {
                swap(l1,l2);
                swap(r1,r2);
            }
            int k=(r1-l1+1),kk=(r2-l2+1),x,xx;
            if((k+kk)%2==1)
            {
                int p=(k+kk+1)/2;
                get(p,x);
                printf("%.1lf\n",(double)a[x]);
            }
            else
            {
                int p=(k+kk)/2;
                int pp=p+1;
                get(p,x),get(pp,xx);
                printf("%.1lf\n",(double)(a[x]+a[xx])/2.0);
            }
        }
    }
    return 0;
}