Codeforces Round #160 (Div. 1) B. Maxim and Restaurant

最新推荐文章于 2020-07-04 11:36:44 发布

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最新推荐文章于 2020-07-04 11:36:44 发布

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分类专栏： DP Codeforces

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Maxim开了一家餐厅，桌子长度为p米。n位客人前来就餐，每位客人的占地面积为ai米。客人按一定顺序排队，当现有客人占用面积加上队列中下一位客人面积大于p时，后面客人无法入座。计算所有可能的客人排队顺序中，平均能进餐厅的客人数量。

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B. Maxim and Restaurant

Maxim has opened his own restaurant! The restaurant has got a huge table, the table's length is p meters.

Maxim has got a dinner party tonight, n guests will come to him. Let's index the guests of Maxim's restaurant from 1 to n. Maxim knows the sizes of all guests that are going to come to him. The i-th guest's size (ai) represents the number of meters the guest is going to take up if he sits at the restaurant table.

Long before the dinner, the guests line up in a queue in front of the restaurant in some order. Then Maxim lets the guests in, one by one. Maxim stops letting the guests in when there is no place at the restaurant table for another guest in the queue. There is no place at the restaurant table for another guest in the queue, if the sum of sizes of all guests in the restaurant plus the size of this guest from the queue is larger than p. In this case, not to offend the guest who has no place at the table, Maxim doesn't let any other guest in the restaurant, even if one of the following guests in the queue would have fit in at the table.

Maxim is now wondering, what is the average number of visitors who have come to the restaurant for all possible n! orders of guests in the queue. Help Maxim, calculate this number.

Input

The first line contains integer n (1 ≤ n ≤ 50) — the number of guests in the restaurant. The next line contains integers a1, a2, ..., an(1 ≤ ai ≤ 50) — the guests' sizes in meters. The third line contains integer p (1 ≤ p ≤ 50) — the table's length in meters.

The numbers in the lines are separated by single spaces.

Output

In a single line print a real number — the answer to the problem. The answer will be considered correct, if the absolute or relative error doesn't exceed 10 - 4.

input
3
1 2 3
3
output
1.3333333333

Note
In the first sample the people will come in the following orders:

(1, 2, 3) — there will be two people in the restaurant;
(1, 3, 2) — there will be one person in the restaurant;
(2, 1, 3) — there will be two people in the restaurant;
(2, 3, 1) — there will be one person in the restaurant;
(3, 1, 2) — there will be one person in the restaurant;
(3, 2, 1) — there will be one person in the restaurant.
In total we get (2 + 1 + 2 + 1 + 1 + 1) / 6 = 8 / 6 = 1.(3)

题意：有一个饭店，很多顾客要进去吃饭每个人有一个宽度a，每个顾客依次进去，如果进去的顾客宽度和超过p了则不能进去。

问再所有的排列组合n!种情况能进去的顾客数的期望是多少。

解法：dp&数学。

如果能够求出所有能进去的情况的人数除以总的排列组合数即为期望的个数。

但肯定不可能用阶乘的时间复杂度，要用dp来解。

用dp[i][j][k]表示前i个人进去j个和为k的组合数。

那么我们只要把所有的f[i][j][k]乘上j与n-j的阶乘以及此时人数(j)就可以表示出人数

但同时为了保证进去j个人之后，不会再进去新的人，所以枚举一个x表示此时x是前j个后的第一个并且x无法进去

然后就可以了。虽然还不是特别理解。dp练的还太少不熟悉，以后再慢慢理解吧。

这里学会一个double存大数的用法，虽然会有精度损失，但如果精度要求不高时还是可以用的。

#include <bits/stdc++.h>
using namespace std;
const int N = 100;
double dp[N][N][N];//dp[i][j][k]表示前i个人进去j个和为k的组合个数
int a[N];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, p, sum = 0;
    cin >> n;
    for(int i  = 1; i <= n; i++)
    {
        cin >> a[i];
        sum += a[i];
    }
    cin >> p;
    double ans = 0;
    double fact[100] = {1,1};
    for(int i = 2; i <= n; i++)
        fact[i] = i * fact[i - 1];
    if(sum <= p)
    {
        cout << n << endl;
        return 0;
    }
    else
    {
        for(int x = 1; x <= n; x++)//枚举最后一个没通过的人
        {
            memset(dp, 0, sizeof(dp));
            for(int i = 0; i <= n; i++)
                dp[i][0][0] = 1;
            for(int i = 1; i <= n; i++)
            {
                for(int j = i; j > 0; j--)
                    for(int k = 0; k <= p; k++)
                        dp[i][j][k] = dp[i - 1][j][k];
                for(int j = i; j > 0; j--)
                    for(int k = 0; k <= p; k++)
                        if(k >= a[i] && x != i)
                            dp[i][j][k] += dp[i][j - 1][k - a[i]];
            }
            for(int k = p; k > p - a[x]; k--)
                for(int j = 0; j <= n; j++)
                    ans += dp[n][j][k] * fact[j] * fact[n - j - 1] * j;
        }
    }
    ans /= fact[n];
    cout << ans << endl;
    return 0;
}