zoukankan      html  css  js  c++  java
  • hdu-5805 NanoApe Loves Sequence(线段树+概率期望)

    题目链接:

    NanoApe Loves Sequence

    Time Limit: 2000/1000 MS (Java/Others)   

     Memory Limit: 262144/131072 K (Java/Others)


    Problem Description
    NanoApe, the Retired Dog, has returned back to prepare for the National Higher Education Entrance Examination!

    In math class, NanoApe picked up sequences once again. He wrote down a sequence with n numbers on the paper and then randomly deleted a number in the sequence. After that, he calculated the maximum absolute value of the difference of each two adjacent remained numbers, denoted as F.

    Now he wants to know the expected value of F, if he deleted each number with equal probability.
     
    Input
    The first line of the input contains an integer T, denoting the number of test cases.

    In each test case, the first line of the input contains an integer n, denoting the length of the original sequence.

    The second line of the input contains n integers A1,A2,...,An, denoting the elements of the sequence.

    1T10, 3n100000, 1Ai109
     
    Output
    For each test case, print a line with one integer, denoting the answer.

    In order to prevent using float number, you should print the answer multiplied by n.
     
    Sample Input
    1 4 1 2 3 4
     
    Sample Output
    6
     
    题意:
     
    问去掉一个数,相邻数的差最大的那个的期望是多少?
     
    思路:
     
    本来很简单的,我傻傻的写了线段树;
     
    AC代码:
     
    /************************************************
    ┆  ┏┓   ┏┓ ┆   
    ┆┏┛┻━━━┛┻┓ ┆
    ┆┃       ┃ ┆
    ┆┃   ━   ┃ ┆
    ┆┃ ┳┛ ┗┳ ┃ ┆
    ┆┃       ┃ ┆ 
    ┆┃   ┻   ┃ ┆
    ┆┗━┓    ┏━┛ ┆
    ┆  ┃    ┃  ┆      
    ┆  ┃    ┗━━━┓ ┆
    ┆  ┃  AC代马   ┣┓┆
    ┆  ┃           ┏┛┆
    ┆  ┗┓┓┏━┳┓┏┛ ┆
    ┆   ┃┫┫ ┃┫┫ ┆
    ┆   ┗┻┛ ┗┻┛ ┆      
    ************************************************ */ 
     
     
    #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=2e3+14;
    const double eps=1e-12;
    
    int a[N],b[N];
    struct node
    {
        int l,r,s;
    }tr[4*N];
    
    void build(int o,int L,int R)
    {
        tr[o].l=L;tr[o].r=R;
        if(L>=R)
        {
            tr[o].s=b[L];
            return ;
        }
        int mid=(L+R)>>1;
        build(2*o,L,mid);
        build(2*o+1,mid+1,R);
        tr[o].s=max(tr[2*o].s,tr[2*o+1].s);
    }
    
    void update(int o,int x,int num)
    {
        if(tr[o].l==tr[o].r&&tr[o].l==x)
        {
            tr[o].s=num;
            return ;
        }
        int mid=(tr[o].l+tr[o].r)>>1;
        if(x<=mid)update(2*o,x,num);
        else update(2*o+1,x,num);
        tr[o].s=max(tr[2*o+1].s,tr[2*o].s);
    }
    int query(int o,int L,int R)
    {
        if(tr[o].l>=L&&tr[o].r<=R)return tr[o].s;
        int mid=(tr[o].r+tr[o].l)>>1;
        if(R<=mid)return query(2*o,L,R);
        else if(L>mid)return query(2*o+1,L,R);
        else return max(query(2*o,L,mid),query(2*o+1,mid+1,R));
    }
    
    int main()
    {      
            int t;
            read(t);
            while(t--)
            {
                int n;
                read(n);
                For(i,1,n)read(a[i]);
                For(i,1,n-1)b[i]=abs(a[i+1]-a[i]);
                build(1,1,n-1);
                LL ans=0;
                For(i,1,n)
                {
                    if(i==1)
                    {
                        update(1,1,0);
                        ans=ans+query(1,1,n-1);
                        update(1,1,b[1]);
                    }
                    else if(i==n)
                    {
                        update(1,n-1,0);
                        ans=ans+query(1,1,n-1);
                        update(1,n-1,b[n-1]);
                    }
                    else 
                    {
                        update(1,i-1,0);
                        update(1,i,abs(a[i+1]-a[i-1]));
                        ans=ans+query(1,1,n-1);
                        update(1,i,b[i]);
                        update(1,i-1,b[i-1]);
                    }
                }
                cout<<ans<<"
    ";
            }      
            return 0;
    }
    

      

     
  • 相关阅读:
    Nginx出现10055错误问题
    MVC报错的坑
    用Docker运行一个简单RazorDemo(一)
    dotnet run
    用Docker 在Centos7上部署Core2.1网站
    VS2015 自己用的插件
    帐户当前被锁定,所以用户 sa 登录失败。系统管理员无法将该帐户解锁
    加密方法汇总
    【备用】网页抓取、
    【转】项目奖金分配
  • 原文地址:https://www.cnblogs.com/zhangchengc919/p/5745163.html
Copyright © 2011-2022 走看看