玲珑杯 Round 19 1152 - Expected value of the expression (dp)

1152 - Expected value of the expressio

You are given an expression: $A_0{O}_1{A}_1{O}_2{A}_2\cdots O_n{A}_n$, where $A_i(0\le i\le n)$ represents number, $O_i(1\le i\le n)$ represents operator. There are three operators, $\&,|,\hat{}$, which means $\text{and},\text{or},\text{xor}$, and they have the same priority.

The $i$-th operator $O_i$ and the numbers $A_i$ disappear with the probability of $p_i$.

Find the expected value of an expression.

扫描二维码关注公众号，回复： 9905886 查看本文章

input：

The first line contains only one integer $n(1\le n\le 1000)$.The second line contains $n+1$ integers $A_i(0\le A_i<2^{20})$.The third line contains $n$ chars $O_i$.The fourth line contains $n$ floats $p_i(0\le p_i\le 1)$.

out:

Output the excepted value of the expression, round to 6 decimal places.

SAMPLE INPUT

21 2 3^ &0.1 0.2

SAMPLE OUTPUT

2.800000

将20位分别开进行考虑计算，dp[i][j] i代表前第几个数运算，在j情况下的概率（j只有两种情况）

#include<stdio.h>
#include<string.h>
#include<math.h>
#define LL long long
const int MAX=1005;
int Ai[MAX];
double p[MAX];
int a[MAX];
char op[MAX][3];
double dp[MAX][3];
int main()
{
 int n;
 int i,j;
 while(scanf("%d",&n)!=EOF)
 {
  for(int i=0;i<=n;i++)scanf("%d",&Ai[i]);
        for(int i=1;i<=n;i++)scanf("%s",op[i]);
        for(int i=1;i<=n;i++)scanf("%lf",&p[i]);
  double output=0;
  for(i=0;i<=20;i++)
  {
   for(int j=0;j<=n;j++)
            {
                if((Ai[j]&(1<<i))>0)   //&优先级低于判断符号
                {
                    a[j]=1;
                }
                else a[j]=0;
            }

   memset(dp,0,sizeof(dp));
   dp[0][a[0]]=1;
     for(int j=1;j<=n;j++)
            {
                dp[j][0]=dp[j-1][0]*p[j];
                dp[j][1]=dp[j-1][1]*p[j];
                if(op[j][0]=='|')
                {
                    if(a[j]==0)
                    {
                        dp[j][0]+=dp[j-1][0]*(1-p[j]);
                        dp[j][1]+=dp[j-1][1]*(1-p[j]);
                    }
                    if(a[j]==1)
                    {
                        dp[j][1]+=dp[j-1][0]*(1-p[j]);
                        dp[j][1]+=dp[j-1][1]*(1-p[j]);
                    }
                }
                if(op[j][0]=='&')
                {
                    if(a[j]==0)
                    {
                        dp[j][0]+=dp[j-1][0]*(1-p[j]);
                        dp[j][0]+=dp[j-1][1]*(1-p[j]);
                    }
                    if(a[j]==1)
                    {
                        dp[j][0]+=dp[j-1][0]*(1-p[j]);
                        dp[j][1]+=dp[j-1][1]*(1-p[j]);
                    }
                }
                if(op[j][0]=='^')
                {
                    if(a[j]==0)
                    {
                        dp[j][0]+=dp[j-1][0]*(1-p[j]);
                        dp[j][1]+=dp[j-1][1]*(1-p[j]);
                    }
                    if(a[j]==1)
                    {
                        dp[j][0]+=dp[j-1][1]*(1-p[j]);
                        dp[j][1]+=dp[j-1][0]*(1-p[j]);
                    }
                }
            }

   output+=(dp[n][1]*(1<<i));
  }
  printf("%.6lf\n",output);
 }
 return 0;
}