ZOJ 3932 Deque and Balls

最新推荐文章于 2022-01-18 21:47:30 发布

Dacc123

最新推荐文章于 2022-01-18 21:47:30 发布

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分类专栏： zoj 动态规划文章标签： dp zoj ZOJ 3932 Deque and B

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探讨一种涉及双端队列的动态规划问题，针对n个标记球放入双端队列的不同方式，计算得到的序列中下降位置的期望值，并通过编程实现。

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There are n balls, where the i-th ball is labeled as p_i. You are going to put n balls into a deque. In the i-th turn, you need to put the i-th ball to the deque. Each ball will be put to both ends of the deque with equal probability.

Let the sequence (x₁, x₂, ..., x_n) be the labels of the balls in the deque from left to right. The beauty of the deque B(x₁, x₂, ..., x_n) is defined as the number of descents in the sequence. For the sequence (x₁, x₂, ..., x_n), a descent is a position i (1 ≤ i < n) with x_i > x_i+1.

You need to find the expected value of B(x₁, x₂, ..., x_n).

Deque is a double-ended queue for which elements can be added to or removed from either the front (head) or the back (tail).

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first line contains an integer n (2 ≤ n ≤ 100000) -- the number of balls. The second line contains n integers: p₁, p₂, ..., p_n (1 ≤ p_i ≤ n).

Output

For each test case, if the expected value is E, you should output E⋅2ⁿ mod (10⁹ + 7).

Sample Input

Sample Output

2
0
第一次遇到求期望的DP题目，每一个数a[i],都可以和前面的数相邻，只要不和自己相同都可以产生新的逆序。所以加上i-1已经
产生的逆序数*2，再加上新产生的逆序数，要减去和相邻是自己相同的数的排列，用一个数组去维护这个排列数

#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdlib.h>
#include <math.h>
#include <stdio.h>

using namespace std;
typedef long long LL;
const LL  mod=1e9+7;
#define MAX 100000
LL dp[MAX+5];
LL p[MAX+5];
LL num[MAX+5];
int n;
int fun()
{
    p[0]=0;p[1]=1;
    for(int i=2;i<=MAX+5;i++)
    {
        p[i]=(p[i-1]*2)%mod;
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    int a;
    fun();
    while(t--)
    {
        scanf("%d",&n);
        memset(dp,0,sizeof(dp));
        memset(num,0,sizeof(num));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a);
            dp[i]=(2*dp[i-1]+p[i-1]-num[a]+mod)%mod;
            if(i==1)
                 num[a]=(num[a]+1)%mod;
            else
                 num[a]=(num[a]+p[i-1])%mod;

        }
        dp[n]=(dp[n]*2)%mod;
        printf("%lld\n",dp[n]);
    }
    return 0;
}