这2dp这么综合起来还是蛮好的。。

看到n范围20马上想状压，发现一起取是个坑丫。。所以状压是正确的。。

然后转移还是先考虑一下正向转移吧。。然而转移的时候子状态转移的概率不相同就很难办了，结果还是得向逆向转移低头orz

然后考虑在当前状态i如果要取到新的卡牌，那么需要算出取到新卡牌的概率t，然后取到新卡牌需要抽取的期望就是1/t了。。然后再加上当前已算出的期望就可以了。。

转移方程看代码方便点。。用方程并不是特别好写。。。



/**
 *　　　　　　　　┏┓　　 　┏┓
 * 　　　　　　　┏┛┗━━━━━━━┛┗━━━┓
 * 　　　　　　　┃　　　　　　　┃ 　
 * 　　　　　　　┃　　　━　　 　┃
 * 　　　　　　　┃　＞　　　＜　┃
 * 　　　　　　　┃　　　　　　　┃
 * 　　　　　　　┃...　⌒　... 　┃
 * 　　　　　　　┃　　　　　　　┃
 * 　　　　　　　┗━┓　　　┏━┛
 * 　　　　　　　　　┃　　　┃　Code is far away from bug with the animal protecting　　　　　　　　　　
 * 　　　　　　　　　┃　　　┃   神兽保佑,代码无bug
 * 　　　　　　　　　┃　　　┃　　　　　　　　　　　
 * 　　　　　　　　　┃　　　┃  　　　　　　
 * 　　　　　　　　　┃　　　┃
 * 　　　　　　　　　┃　　　┃　　　　　　　　　　　
 * 　　　　　　　　　┃　　　┗━━━┓
 * 　　　　　　　　　┃　　　　　　　┣┓
 * 　　　　　　　　　┃　　　　　　　┏┛
 * 　　　　　　　　　┗┓┓┏━┳┓┏┛
 * 　　　　　　　　　　┃┫┫　┃┫┫
 * 　　　　　　　　　　┗┻┛　┗┻┛
 */ 
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define ll long long
#define eps 1e-12
#define succ(x) (1<<x)
#define lowbit(x) (x&(-x))
#define sqr(x) ((x)*(x))
#define mid (x+y>>1)
#define NM 2000050
#define nm 7000000
#define pi 3.1415926535897931
using namespace std;
const ll inf=1000000;
ll read(){
    ll x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    return f*x;
}
 






int n,tot;
double a[25],d[NM];
int main(){
    freopen("data.in","r",stdin);
    while(~scanf("%d",&n)){
        mem(d);
        inc(i,1,n)scanf("%lf",a+i);
        tot=succ(n)-1;
        dec(i,tot-1,0){
            double t=0;
            inc(j,1,n)if(!(i&succ(j-1)))t+=a[j];
            inc(j,1,n)if(!(i&succ(j-1)))
                d[i]+=d[i|succ(j-1)]*a[j]/t;
            d[i]+=1/t;
        }
        printf("%lf\n",d[0]);
    }
    return 0;
}








Card Collector


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5118    Accepted Submission(s): 2605
Special Judge


Problem Description


In your childhood, do you crazy for collecting the beautiful cards in the snacks? They said that, for example, if you collect all the 108 people in the famous novel Water Margin, you will win an amazing award.

As a smart boy, you notice that to win the award, you must buy much more snacks than it seems to be. To convince your friends not to waste money any more, you should find the expected number of snacks one should buy to collect a full suit of cards.

Input


The first line of each test case contains one integer N (1 <= N <= 20), indicating the number of different cards you need the collect. The second line contains N numbers p1, p2, ..., pN, (p1 + p2 + ... + pN <= 1), indicating the possibility of each card to appear in a bag of snacks.

Note there is at most one card in a bag of snacks. And it is possible that there is nothing in the bag.


 



Output


Output one number for each test case, indicating the expected number of bags to buy to collect all the N different cards.

You will get accepted if the difference between your answer and the standard answer is no more that 10^-4.

Sample Input


1 0.1 2 0.1 0.4


Sample Output


10.000 10.500


Source


​​2012 Multi-University Training Contest 4 ​​


 



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