HDU 5857 Median

因为原序列是排列好了的，那么只要看一下给出的两个区间相交的情况，然后分类讨论一下，O(1)输出。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<iostream>
using namespace std;
typedef long long LL;
const double pi=acos(-1.0),eps=1e-8;
void File()
{
    freopen("D:\\in.txt","r",stdin);
    freopen("D:\\out.txt","w",stdout);
}
template <class T>
inline void read(T &x)
{
    char c = getchar(); x = 0;while(!isdigit(c)) c = getchar();
    while(isdigit(c)) { x = x * 10 + c - ‘0‘; c = getchar();  }
}

const int maxn=100010;
int T,n,m,a[maxn];

int get(int l1,int r1,int l2,int r2 ,int x)
{
    if(l2>r1)
    {
        if(r1-l1+1>=x) return a[l1+x-1];
        else return a[l2+x-(r1-l1+1)-1];
    }

    else if(r2<=r1)
    {
        if(l2-l1>=x) return a[l1+x-1];
        else if(2*(r2-l2+1)+l2-l1>=x) return a[l2+(x-(l2-l1)-1)/2];
        else return a[r2+1+x-(2*(r2-l2+1)+l2-l1)-1];
    }

    else
    {
        if(l2-l1>=x) return a[l1+x-1];
        else if(2*(r1-l2+1)+l2-l1>=x) return a[l2+(x-(l2-l1)-1)/2];
        else return a[r1+1+x-(2*(r1-l2+1)+l2-l1)-1];
    }
}

int main()
{
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        for(int i=1;i<=m;i++)
        {
            int l1,r1,l2,r2;
            scanf("%d%d%d%d",&l1,&r1,&l2,&r2);
            if(l1>l2) { swap(l1,l2); swap(r1,r2); }
            if((r1-l1+1+r2-l2+1)%2==1)
            {
                printf("%.1lf\n",1.0*get(l1,r1,l2,r2,(r1-l1+1+r2-l2+1)/2+1));
            }
            else
            {
                double num1=1.0*get(l1,r1,l2,r2,(r1-l1+1+r2-l2+1)/2);
                double num2=1.0*get(l1,r1,l2,r2,(r1-l1+1+r2-l2+1)/2+1);
                printf("%.1lf\n",1.0*(num1+num2)/2.0);
            }
        }
    }
    return 0;
}