首页 > 代码库 > 概率dp ZOJ 3640

概率dp ZOJ 3640

Help Me Escape
Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu
Submit Status Practice ZOJ 3640
Appoint description: 

Description

Background

    If thou doest well, shalt thou not be accepted? and if thou doest not well, sin lieth at the door. And unto thee shall be his desire, and thou shalt rule over him. 
    And Cain talked with Abel his brother: and it came to pass, when they were in the field, that Cain rose up against Abel his brother, and slew him. 
    And the LORD said unto Cain, Where is Abel thy brother? And he said, I know not: Am I my brother‘s keeper? 
    And he said, What hast thou done? the voice of thy brother‘s blood crieth unto me from the ground. 
    And now art thou cursed from the earth, which hath opened her mouth to receive thy brother‘s blood from thy hand; 
    When thou tillest the ground, it shall not henceforth yield unto thee her strength; a fugitive and a vagabond shalt thou be in the earth.

—— Bible Chapter 4

Now Cain is unexpectedly trapped in a cave with N paths. Due to LORD‘s punishment, all the paths are zigzag and dangerous. The difficulty of the ith path is ci.

Then we define f as the fighting capacity of Cain. Every day, Cain will be sent to one of the N paths randomly.

Suppose Cain is in front of the ith path. He can successfully take ti days to escape from the cave as long as his fighting capacity f is larger than ci. Otherwise, he has to keep trying day after day. However, if Cain failed to escape, his fighting capacity would increase cias the result of actual combat. (A kindly reminder: Cain will never died.)

As for ti, we can easily draw a conclusion that ti is closely related to ci. Let‘s use the following function to describe their relationship:

After D days, Cain finally escapes from the cave. Please output the expectation of D.

Input

The input consists of several cases. In each case, two positive integers N and f (n ≤ 100, f ≤ 10000) are given in the first line. The second line includes N positive integers ci (ci ≤ 10000, 1 ≤ i ≤ N)

Output

For each case, you should output the expectation(3 digits after the decimal point).

Sample Input

3 1
1 2 3

Sample Output

6.889

/*************************************************************************
    > File Name: t.cpp
    > Author: acvcla
    > Mail: acvcla@gmail.com 
    > Created Time: 2014年10月21日 星期二 21时33分55秒
 ************************************************************************/
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<cstring>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<cstdlib>
#include<ctime>
#include<set>
#include<math.h>
using namespace std;
typedef long long LL;
const int maxn = 20000 + 10;
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define pb push_back
int n,c[105],f;
double dp[maxn];
double d(int f){
	if(dp[f]>0)return dp[f];
	dp[f]=0;
	for(int i=1;i<=n;i++){
		if(f>c[i]){
			int t=(1+sqrt(5))*c[i]*c[i]/2;
			dp[f]+=(double)t/n;
		}else{
			dp[f]+=(1+d(f+c[i]))/n;
		}
	}
	return dp[f];
}
int main(int argc, char const *argv[])
{
	while(~scanf("%d%d",&n,&f)){
		memset(dp,0,sizeof dp);
		for(int i=1;i<=n;i++)scanf("%d",c+i);
		printf("%.3f\n",d(f));
	}
	return 0;
}



概率dp ZOJ 3640