hdu5857 Median（模拟）

Median

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2049    Accepted Submission(s): 506

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

1 4 2 1 2 3 4 1 2 2 4 1 1 2 2

 

 

Sample Output

2.0 1.5

 

 

Author

BUPT

 

 

Source

2016 Multi-University Training Contest 10

 

 

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题意：给出一个有序的数列.
求由 A[l1]~A[r1] 与 A[l2]~A[r2] 组成的新序列的中位数.

代码：

#include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std;
#define MAX 100005
int l1,l2,r2,r1;
int n,m;
int a[MAX];
int solve(int len)
{
    if(r1<=l2)
    {
        if(r1-l1+1>=len) return a[l1+len-1];
        else{
            len-=r1-l1+1;
            return a[l2+len-1];
        }
    }
    else{
            if(l2-l1>=len)
            {
                return a[l1+len-1];
            }
       else  if((l2-l1+2*(r1-l2+1))<len){
            len-=(l2-l1+2*(r1-l2+1));//注意这里的括号没加就变成len-l2+l1-2*(r1-l2+1)
            return a[r1+len];
        }
        else{
            len-=l2-l1;
            if(len&1)
            {
                return a[l2+len/2];
            }
            else{
                return a[l2+len/2-1];
            }
        }
    }
}
int main()
{
    int T;
   scanf("%d",&T);
   while(T--)
   {
       scanf("%d %d",&n,&m);
       for(int i=1;i<=n;i++)
       {
           scanf("%d",&a[i]);
       }
       while(m--){
       scanf("%d %d %d %d",&l1,&r1,&l2,&r2);
       if(l1>l2)swap(l1,l2);
       if(r1>r2)swap(r1,r2);
       int len=(r2-l2+1)+(r1-l1+1);
       if(len&1)
       {
           printf("%.1f\n",1.0*solve(len/2+1));
       }
       else printf("%.1f\n",0.5*solve(len/2)+0.5*solve(len/2+1));
       }
   }
}