A - Anniversary Firework
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit Status Practice URAL
1776
Appoint description:
System Crawler (2016-05-06)
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
| input | output |
|---|---|
5 |
26.66666666666 |
Notes
First, the rockets with numbers 1 and 5 are launched. 10 seconds later the rocket 3 is launched with probability 1/3; in that case, 10 more seconds later the rockets 2 and 4 are launched, and the firework is over after 20 seconds. In case the rocket 2 or
rocket 4 is launched in the second salvo (this happens with probability 2/3), the firework is over after 30 seconds.
题目的意思给你n个火箭排成一排
一开始点燃第一个和最后一个火箭
然后每次只能在点燃过的火箭中的火箭
每两次点燃火箭的间隔时间为10s求
点燃n个火箭等待时间的期望
设dp【i，j】表示i个火箭等待了j次
那么求花费j次的概率为 dp[【i，j】=dp【i，j-1】
然后就枚举长度和次数
ACcode：
#include <cstdio>
#include <cstring>
#include <iostream>
#define maxn 404
using namespace std;
double dp[maxn][maxn];
int main(){
int n;
while(~scanf("%d",&n)){
n-=2;
for(int i=0;i<=n;++i)for(int j=i;j<=n;++j)dp[i][j]=1.0;
for(int i=1;i<=n;i++){
double e=1.0/i;
for(int j=2;j<i; j++){
dp[i][j]=dp[i][j-1];
for(int k=1;k<=i;k++){
int l=k-1,r=i-k;
double p1=dp[l][j-1],p2=dp[r][j-1],p3=dp[l][j-2],p4=dp[r][j-2];
dp[i][j]+=e*(p1*p2-p3*p4);
}
}
}
double ans = 0;
for(int i = 1; i <= n; i++){
ans += (dp[n][i]-dp[n][i-1])*i*10;
}
printf("%.11lf\n",ans);
}
return 0;
}