[codeforces167B]Wizards and Huge Prize

B. Wizards and Huge Prize

time limit per test:

2 seconds

memory limit per test:

256 megabytes

input:

standard input

output:standard output

One must train much to do well on wizardry contests. So, there are numerous wizardry schools and magic fees.

One of such magic schools consists of n tours. A winner of each tour gets a huge prize. The school is organised quite far away, so one will have to take all the prizes home in one go. And the bags that you've brought with you have space for no more than k huge prizes.

Besides the fact that you want to take all the prizes home, you also want to perform well. You will consider your performance good if you win at least l tours.

In fact, years of organizing contests proved to the organizers that transporting huge prizes is an issue for the participants. Alas, no one has ever invented a spell that would shrink the prizes... So, here's the solution: for some tours the winner gets a bag instead of a huge prize. Each bag is characterized by number ai — the number of huge prizes that will fit into it.

You already know the subject of all tours, so you can estimate the probability pi of winning the i-th tour. You cannot skip the tour under any circumstances.

Find the probability that you will perform well on the contest and will be able to take all won prizes home (that is, that you will be able to fit all the huge prizes that you won into the bags that you either won or brought from home).

Input

The first line contains three integers n, l, k (1 ≤ n ≤ 200, 0 ≤ l, k ≤ 200) — the number of tours, the minimum number of tours to win, and the number of prizes that you can fit in the bags brought from home, correspondingly.

The second line contains n space-separated integers, pi (0 ≤ pi ≤ 100) — the probability to win the i-th tour, in percents.

The third line contains n space-separated integers, ai (1 ≤ ai ≤ 200) — the capacity of the bag that will be awarded to you for winning the i-th tour, or else -1, if the prize for the i-th tour is a huge prize and not a bag.

Output

Print a single real number — the answer to the problem. The answer will be accepted if the absolute or relative error does not exceed10 - 6.

Examples

input

3 1 0
10 20 30
-1 -1 2

output

0.300000000000

input

1 1 1
100
123

output

1.000000000000

Note

In the first sample we need either win no tour or win the third one. If we win nothing we wouldn't perform well. So, we must to win the third tour. Other conditions will be satisfied in this case. Probability of wining the third tour is 0.3.

In the second sample we win the only tour with probability 1.0, and go back home with bag for it.

题解

这大概是我第一道不是图上的概率

刚开始的时候都没有什么思路......果然dp还是要多刷题

我们设f[i][j][k]为"前i天赢了j场,剩余空间为k"的概率

通过十分艰苦地读题,不难发现,k在超过200后就没有什么用了,所以k只需要枚举0~200

但我们还发现这些比赛是无序的,也就是说我们可以先去拿后面的包,再去拿前面的奖,所以我们必须把它变成无序的.

考虑把k都加上200(诡异的思路......),这样前面的奖品也可以先选上(+200后是正的),再去后面选包,这样就可以随意处理了

这样最后在200<=k<=400,l<=j<=n的范围内枚举所有f[n][j][k]即可(现在的200意味着原来的0,小于200意味着剩余空间为负值)

代码见下:

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int N=210;
double p[N],k[N];
int n,l,s,b[N];
double f[N][N][2*N];
int main()
{
    scanf("%d%d%d",&n,&l,&s);
    for(int i=1;i<=n;i++)
        scanf("%lf",&p[i]),p[i]/=100,k[i]=1.0-p[i];
    for(int i=1;i<=n;i++)
        {scanf("%d",&b[i]);}
    f[0][0][s+200]=1;
    for(int i=0;i<n;i++)
        for(int j=0;j<=i;j++)
            for(int v=0;v<=400;v++)
            {
                int t=min(v+b[i+1],400);
                f[i+1][j][v]+=f[i][j][v]*k[i+1];
                if(t>=0)
                    f[i+1][j+1][t]+=f[i][j][v]*p[i+1];
            }
    double ans=0.0;
    for(int v=200;v<=400;v++)
        for(int j=l;j<=n;j++)
            ans+=f[n][j][v];
    printf("%.12lf",ans);    
}