HDU5857 Median

Median

                                                                      Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
                                                                                             Total Submission(s): 1509    Accepted Submission(s): 396

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

14 21 2 3 41 22 41 12 2

 

Sample Output

2.01.5

 

Author

BUPT

 

Source

2016 Multi-University Training Contest 10

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题目的意思是给出一个序列，要求查询两个区间，将两个区间的数合并后求中位数

思路：分类讨论大法好  注意数据会爆int

#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 longconst 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;}