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HDU 4336 Card Collector

状态压缩,期望,$dp$。

设$dp[s]$表示状态$s$中的牌已经获得了需要的期望包数,$dp[0]$为答案。$dp[2^n-1]=0$,倒着推一遍就可以得到答案了。保留三位小数$WA$了,保留四位对了。

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<iostream>using namespace std;typedef long long LL;const double pi=acos(-1.0),eps=1e-6;void File(){    freopen("D:\\in.txt","r",stdin);    freopen("D:\\out.txt","w",stdout);}template <class T>inline void read(T &x){    char c = getchar();    x = 0;    while(!isdigit(c)) c = getchar();    while(isdigit(c))    {        x = x * 10 + c - 0;        c = getchar();    }}int n;double p[30];double dp[(1<<20)+10],P;int main(){    while(~scanf("%d",&n))    {        P=0;        for(int i=0;i<n;i++)        {            scanf("%lf",&p[i]);            P=P+p[i];        }        for(int i=0;i<(1<<n);i++) dp[i]=0;        for(int i=(1<<n)-2;i>=0;i--)        {            double sum1=0,sum2=0;            for(int j=0;j<n;j++)            {                if((1<<j)&i) sum1=sum1+p[j];                else sum2=sum2+dp[(1<<j)+i]*p[j];            }            sum2=sum2+1.0;            sum1=sum1+1.0-P;            dp[i]=sum2/(1.0-sum1);        }        printf("%.4f\n",dp[0]);    }    return 0;}

 

HDU 4336 Card Collector