HDU 4649 Professor Tian (2013多校联合5 1007)

http://acm.hdu.edu.cn/showproblem.php?pid=4649

 

Professor Tian

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 76    Accepted Submission(s): 48

Problem Description

 

Timer took the Probability and Mathematical Statistics course in the 2012, But his bad attendance angered Professor Tian who is in charge of the course. Therefore, Professor Tian decided to let Timer face a hard probability problem, and announced that if he fail to slove the problem there would be no way for Timer to pass the final exam. 
As a result , Timer passed. 
Now, you, the bad guy, also angered the Professor Tian when September Ends. You have to faced the problem too. The problem comes that there is an expression and you should calculate the excepted value of it. And the operators it may contains are '&' (and),'|'(or) and '^'(xor) which are all bit operators. For example: 7&3=3, 5&2=0, 2|5=7, 4|10=14, 6^5=3, 3^4=7.
Professor Tian declares that each operator O_i with its coming number A_i+1 may disappear, and the probability that it happens is P_i (0<i<=n). 

 

 

Input

 

The input contains several test cases. For each test case, there is a integer n (0<n<=200) in the first line.In the second line, there are n+1 integers, stand for {A_i}. The next line contains n operators ,stand for {O_i}. The forth line contains {P_i}. 
A_i will be less than 2²⁰, 0<=P_i<=1.

 

 

Output

 

For each text case, you should print the number of text case in the first line.Then output the excepted value of the expression, round to 6 decimal places.

 

比较简单的一道DP问题，我们设dp[i][j]表示前i个数计算完之后第j位为1的概率（二进制），那最后的答案就是

dp[n][0]*2^0+dp[n][1]*2^1+dp[n][2]*2^2+......dp[n][20]*2^20.（这里的^为乘方运算）

对于dp[i][j]，设第i个符号消失的概率为pi,第（i+1）个数的第j位为di，则分三种情况考虑：

1.第i个符号为&，这时若di为1，则dp[i][j]=dp[i-1][j],否则，dp[i][j]=pi*dp[i-1][j];

2.第i个符号为|，这时若di为1，则 dp[i][j]=pi*dp[i-1][j]+1-pi，否则，dp[i][j]=dp[i-1][j];

3.第i个符号位^,这时若di为1，则dp[i][j]=dp[i-1][j]*pi+(1-pi)*(1-dp[i-1][j]);，否则，dp[i][j]=dp[i-1][j];

最后统计答案即可。注意初始化，dp[0][j]=(a[0]>>j)&1(0<=j<=20)。代码如下：

 

#include <iostream>#include <string.h>#include <stdio.h>#include <algorithm>using namespace std;double dp[210][25];char op[210];double pi[210];int a[210];int main(){    //freopen("dd.txt","r",stdin);    int n,i,j,T=0;    while(scanf("%d",&n)!=EOF)    {        printf("Case %d:\n",++T);        memset(dp,0,sizeof(dp));        for(i=0;i<=n;i++)        scanf("%d",&a[i]);        for(i=1;i<=n;i++)        cin>>op[i];        for(i=1;i<=n;i++)        scanf("%lf",&pi[i]);        for(i=0;i<=20;i++)        {            dp[0][i]=(double)((a[0]>>i)&1);        }        for(i=1;i<=n;i++)        {            for(j=0;j<=20;j++)            {                int tmp=(1&(a[i]>>j));                if(op[i]=='&')                {                    if(tmp)                    {                        dp[i][j]=dp[i-1][j];                    }                    else                    {                        dp[i][j]=pi[i]*dp[i-1][j];                    }                }                else if(op[i]=='|')                {                    if(tmp)                    {                        dp[i][j]=pi[i]*dp[i-1][j]+1-pi[i];                    }                    else                    {                        dp[i][j]=dp[i-1][j];                    }                }                else if(op[i]=='^')                {                    if(tmp)                    {                        dp[i][j]=dp[i-1][j]*pi[i]+(1-pi[i])*(1-dp[i-1][j]);                    }                    else                    {                        dp[i][j]=dp[i-1][j];                    }                }            }        }        double ans=0;        for(i=0;i<=20;i++)        {            ans+=(double)(1<<i)*dp[n][i];        }        printf("%.6lf\n",ans);    }    return 0;}