Codeforces 262D Maxim and Restaurant

最新推荐文章于 2019-08-21 10:27:03 发布

steveyg

最新推荐文章于 2019-08-21 10:27:03 发布

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分类专栏： Codeforces

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本文探讨了一种餐厅顾客排队就餐的数学模型问题，通过计算不同顾客顺序进入长度固定的餐厅就餐时，平均能接纳的顾客数量。该问题涉及排列组合及概率计算，并提供了一个具体的算法实现案例。

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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.

Sample test(s)

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).

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <sstream>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <ctime>
#define MAXN 1000  
#define offset 10000  
#define eps 1e-11  
#define PI acos(-1.0)//3.14159265358979323846  
#define exp 2.718281828
#define mod 1000000007
using namespace std;
typedef long long LL;
double fac[55],f[55][55][55];
int a[55];
int main(){

	fac[0] = 1;
	for(int i = 1;i<=54;i++){
		fac[i] = fac[i-1] * i;
	}
	int n;
	scanf("%d",&n);
	int sum = 0;
	for(int i = 1;i<=n;i++){
		scanf("%d",&a[i]);
		sum += a[i];
	}
	int p;
	scanf("%d",&p);
	if(sum <= p){
		printf("%d\n",n);
	}else{
		for(int i = 1;i<=n;i++){
			f[0][0][i] = 1;
		}
		for(int i = 1;i<=n;i++){
			for(int j = 1;j<=n;j++){
				if(i != j){
					for(int k = n;k>=1;k--){
						for(int x = p;x>=0&&x>=a[i];x--){
							f[x][k][j] += f[x-a[i]][k-1][j];
						}
					}
				}
			}
		}
		double ans = 0;
		for(int i = 1;i<=n;i++){
			if(a[i] > p){
				a[i] = p+1; 
			}
			
			for(int x = p+1-a[i];x<=p;x++){
				for(int k = 1;k<=n;k++){
					ans += f[x][k][i]*fac[k]*fac[n-k-1]*k;
				}
			}
		}
	//		printf("%.7lf\n",ans);
		ans /= fac[n];
		printf("%.7lf\n",ans);

	}
	return 0;
}