zoukankan      html  css  js  c++  java
  • hdu-5857 Median(水题)

    题目链接:

    Median

    Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 21    Accepted Submission(s): 4


    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
     
    题意:
     
    给以一个排好序的序列,m个询问,每个询问给两个区间,问用这两个区间的这些数组成的新序列的中位数是多少;
     
    思路:
     
    把区间分成三部分,其中一个是相交部分,然后找中位数就好了;
     
    AC代码:
     
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <cmath>
    #include <bits/stdc++.h>
    #include <stack>
    #include <map>
     
    using namespace std;
     
    #define For(i,j,n) for(int i=j;i<=n;i++)
    #define mst(ss,b) memset(ss,b,sizeof(ss));
     
    typedef  long long LL;
     
    template<class T> void read(T&num) {
        char CH; bool F=false;
        for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
        for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
        F && (num=-num);
    }
    int stk[70], tp;
    template<class T> inline void print(T p) {
        if(!p) { puts("0"); return; }
        while(p) stk[++ tp] = p%10, p/=10;
        while(tp) putchar(stk[tp--] + '0');
        putchar('
    ');
    }
     
    const LL mod=1e9+7;
    const double PI=acos(-1.0);
    const int inf=1e9;
    const int N=1e5+10;
    const int maxn=(1<<20)+14;
    const double eps=1e-12;
    
    int n,m,a[N],l1,r1,l2,r2,l3,r3;
    
    int solve(int pos)
    {
        if(r1>=l1)
        {
            if(r1-l1+1>=pos)return a[l1+pos-1];
            else pos-=(r1-l1+1);
        }
        if(r3>=l3)
        {
            if(2*(r3-l3+1)>=pos)
            {
                if(pos&1)pos=pos/2+1;
                else pos=pos/2;
                return a[l3+pos-1];
            }
            else pos-=2*(r3-l3+1);
        }
        if(r2>=l2)
        {
            if(r2-l2+1>=pos)return a[l2+pos-1];
            else pos-=(r2-l2+1);
        }
    }
    
    int main()
    {
        int t;
        read(t);
        while(t--)
        {
            read(n);read(m);
            For(i,1,n)read(a[i]);
            while(m--)
            {
                read(l1);read(r1);
                read(l2);read(r2);
                int num=(r1-l1+1)+(r2-l2+1);
                if(l1>l2)swap(l1,l2);
                if(r1>r2)swap(r1,r2);
                if(r1>=l2)
                {
                    l3=l2,r3=r1;
                    r1=l3-1;
                    l2=r3+1;
                }
                else r3=-1,l3=0;
                if(num&1)printf("%.1lf
    ",solve(num/2+1)*1.0);
                else printf("%.1lf
    ",solve(num/2)*0.5+solve(num/2+1)*0.5);
            }
        }
        return 0;
    }
    

      

  • 相关阅读:
    归并两路有序链表
    [转]两种高性能I/O设计模式(Reactor/Proactor)的比较
    linux 静态库使用经验
    系统性能调优经验
    编译-O 选项对性能提升作用
    [转]Linux shell中的那些小把戏
    shell函数传递带空格的参数
    标题清洗引发的算法(两个字符串的最长公共子串)
    正则表达式之Matcher类中group方法
    ConcurrentHashMap JDK 1.6 源码分析
  • 原文地址:https://www.cnblogs.com/zhangchengc919/p/5784658.html
Copyright © 2011-2022 走看看