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=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } //************************************************************************************** int v[500][500]; double dp[500][500]; double dfs(int x,int y) { if(x==0)return 1.0; if(y==0)return 0.0; if(v[x][y]) return dp[x][y]; double p = 1.0/(x*1.0); double ans=0; for(int i=1;i<=x;i++) ans+=p*(dfs(i-1,y-1)*dfs(x-i,y-1)); v[x][y]=1; dp[x][y]=ans; return ans; } int main() { int n; cin>>n; double ans=0; for(int i=1;i<=n-2;i++) ans+=(dfs(n-2,i)-dfs(n-2,i-1))*(1.0*10*i); printf("%.10lf ",ans); }