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.

1≤T≤10, 3≤n≤100000, 1≤Ai≤109

 

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('\n');
}

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<<"\n";
        }
        return 0;
}