概率dp+状态压缩HDU4336
2014-10-25 21:53
232 查看
Card Collector
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d
& %I64u
Submit Status Practice HDU
4336
Appoint description:
System Crawler (2014-10-23)
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
/*************************************************************************
> File Name: t.cpp
> Author: acvcla
> Mail: acvcla@gmail.com
> Created Time: 2014年10月21日 星期二 21时33分55秒
************************************************************************/
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<cstring>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<cstdlib>
#include<ctime>
#include<set>
#include<math.h>
using namespace std;
typedef long long LL;
const int maxn = 1e5 + 10;
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define pb push_back
double dp[1<<20],p[1<<20],A[200];
void Init(int n){
memset(p,0,sizeof p);
memset(dp,0,sizeof dp);
for(int i=(1<<n)-1;i>0;i--){
double t=0;
for(int j=0;j<n;j++)if(1<<j&i)
{
p[i]+=A[j];
}
}
}
int n;
int main(int argc, char const *argv[])
{
while(~scanf("%d",&n)){
double s=1;
for(int i=0;i<n;i++){
scanf("%lf",A+i);
s-=A[i];
}
Init(n);
for(int i=(1<<n)-2;i>=0;i--){
double t=0;
double pi=p[i]+s;
for(int j=0;j<n;j++){
if((1<<j)&i)continue;
else{
int temp=i|(1<<j);
t+=dp[temp]*A[j];
}
}
dp[i]=(t+1)/(1-pi);
}
printf("%.4f\n",dp[0]);
}
return 0;
}
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d
& %I64u
Submit Status Practice HDU
4336
Appoint description:
System Crawler (2014-10-23)
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
/*************************************************************************
> File Name: t.cpp
> Author: acvcla
> Mail: acvcla@gmail.com
> Created Time: 2014年10月21日 星期二 21时33分55秒
************************************************************************/
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<cstring>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<cstdlib>
#include<ctime>
#include<set>
#include<math.h>
using namespace std;
typedef long long LL;
const int maxn = 1e5 + 10;
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define pb push_back
double dp[1<<20],p[1<<20],A[200];
void Init(int n){
memset(p,0,sizeof p);
memset(dp,0,sizeof dp);
for(int i=(1<<n)-1;i>0;i--){
double t=0;
for(int j=0;j<n;j++)if(1<<j&i)
{
p[i]+=A[j];
}
}
}
int n;
int main(int argc, char const *argv[])
{
while(~scanf("%d",&n)){
double s=1;
for(int i=0;i<n;i++){
scanf("%lf",A+i);
s-=A[i];
}
Init(n);
for(int i=(1<<n)-2;i>=0;i--){
double t=0;
double pi=p[i]+s;
for(int j=0;j<n;j++){
if((1<<j)&i)continue;
else{
int temp=i|(1<<j);
t+=dp[temp]*A[j];
}
}
dp[i]=(t+1)/(1-pi);
}
printf("%.4f\n",dp[0]);
}
return 0;
}
相关文章推荐
- hdu4336(概率DP+状态压缩)
- hdu4336(概率DP+状态压缩)
- codeforces 424E Colored Jenga (状态压缩,概率dp用hash记忆优化搜索)
- 【BZOJ 3925】[Zjoi2015]地震后的幻想乡 期望概率dp+状态压缩+图论知识+组合数学
- HDU 4336 Card Collector [状态压缩概率DP]
- hdu4336 Card Collector 状态压缩dp
- hdu 4649 Professor Tian 反状态压缩+概率DP
- Hdu 4336 Card Collector 概率DP+状态压缩
- 【BZOJ-1076】奖励关 概率与期望 + 状态压缩DP
- codeforces 16E Fish(状态压缩概率DP)
- ZOJ 3502 Contest <状态压缩 概率 DP>
- HDU 4336 Card Collector 概率dp 状态压缩
- hdu 4336 Card Collector(状态压缩概率dp)
- HDU 4336 Card Collector 状态压缩+概率DP
- HDU 4336 Card Collector 概率dp 状态压缩| 容斥原理
- hdu 4336 全期望公式+状态压缩+概率dp
- HDU 4336 Card Collector(概率DP,状态压缩)
- [HDU 4336]Card Collection[状态压缩DP][概率DP][容斥原理]
- hdu 4336 Card Collector 概率DP 状态压缩DP
- HDOJ4336Card Collector【概率dp求期望+状态压缩】