C - Anniversary Firework
Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87643#problem/B
Description
Denis has to prepare the Ural State University 90th anniversary firework. He bought n rockets and started to think of the way he should launch them. After a pair of sleepless nights he invented the following algorithm.
All n rockets are placed on the surface in a single line. The interval between two consecutive salvos is ten seconds. The leftmost and the rightmost rocket are launched in the first salvo. After i salvos are fired, all non-empty segments between two neighboring launched rockets are considered. One rocket is chosen randomly and uniformly at each of these segments. All chosen rockets are launched in the (i + 1)-st salvo. Algorithm runs until all rockets are launched.
Calculate the average duration in seconds of such a firework.
Input
The only input line contains an integer n (3 ≤ n ≤ 400) , which is the number of rockets bought by Denis.
Output
Output the expected duration of the firework in seconds, with absolute or relative error not exceeding 10 −6.
Sample Input
5
Sample Output
26.66666666666
HINT
题意
有n个火箭,每次你都会在已经放了火箭的区间内随机选择一个放火箭
然后问你把火箭全部放完的期望时间是多少
题解:
首先,期望取最大值这个是错误的
这个dp[i][j]应该表示长度为i的,放j个火箭的概率是多少
直接dfs解决就好了
代码:
//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 5000
#define mod 10007
#define eps 1e-9
int Num;
//const int inf=0x7fffffff; //нчоч╢С
const int inf=0x3f3f3f3f;
inline ll read()
{
ll x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
//************************************************************************************** int v[][];
double dp[][];
double dfs(int x,int y)
{
if(x==)return 1.0;
if(y==)return 0.0;
if(v[x][y])
return dp[x][y];
double p = 1.0/(x*1.0);
double ans=;
for(int i=;i<=x;i++)
ans+=p*(dfs(i-,y-)*dfs(x-i,y-));
v[x][y]=;
dp[x][y]=ans;
return ans;
}
int main()
{
int n;
cin>>n;
double ans=;
for(int i=;i<=n-;i++)
ans+=(dfs(n-,i)-dfs(n-,i-))*(1.0**i);
printf("%.10lf\n",ans);
}