玲珑学院oj 1152 概率dp

1152 - Expected value of the expression

Time Limit：2s Memory Limit：128MByte

Submissions：128Solved：63

DESCRIPTION

You are given an expression: A0O1A1O2A2⋯OnAnA0O1A1O2A2⋯OnAn, where Ai(0≤i≤n)Ai(0≤i≤n) represents number, Oi(1≤i≤n)Oi(1≤i≤n) represents operator. There are three operators, &,|,^&,|,^, which means and,or,xorand,or,xor, and they have the same priority.

The ii-th operator OiOi and the numbers AiAi disappear with the probability of pipi.

Find the expected value of an expression.

INPUT

The first line contains only one integer n(1≤n≤1000)n(1≤n≤1000). The second line contains n+1n+1 integers Ai(0≤Ai<220)Ai(0≤Ai<220). The third line contains nn chars OiOi. The fourth line contains nn floats pi(0≤pi≤1)pi(0≤pi≤1).

OUTPUT

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

SAMPLE INPUT

2

1 2 3

^ &

0.1 0.2

SAMPLE OUTPUT

2.800000

HINT

Probability = 0.1 * 0.2 Value = 1 Probability = 0.1 * 0.8 Value = 1 & 3 = 1 Probability = 0.9 * 0.2 Value = 1 ^ 2 = 3 Probability = 0.9 * 0.8 Value = 1 ^ 2 & 3 = 3 Expected Value = 0.1 * 0.2 * 1 + 0.1 * 0.8 * 1 + 0.9 * 0.2 * 3 + 0.9 * 0.8 * 3 = 2.80000

SOLUTION

“玲珑杯”ACM比赛 Round #19

题意：给你n+1个数，n个位运算符，第一个数到第n个数和对应的n个运算符一起消失的概率为p[i]，问你运算结果的期望。

题解：dp[i][j][k]   前i个数 a[i]二进制第j位置填k的概率 具体看代码中的转移方程

 #pragma comment(linker, "/STACK:102400000,102400000")
 #include <bits/stdc++.h>
 #include <cstdlib>
 #include <cstdio>
 #include <iostream>
 #include <cstdlib>
 #include <cstring>
 #include <algorithm>
 #include <cmath>
 #include <cctype>
 #include <map>
 #include <set>
 #include <queue>
 #include <bitset>
 #include <string>
 #include <complex>
 #define ll long long
 #define mod 1000000007
 using namespace std;
 int n;
 char s[];
 int a[];
 char o[];
 double p[];
 double dp[][][];
 int main()
 {
     scanf("%d",&n);
     for(int i=;i<=n;i++){
         scanf("%d",&a[i]);
     }
     getchar();
     gets(s);
     int len=strlen(s);
     int res=;
     for(int i=;i<len;i++){
         if(s[i]!=' '){
          o[res++]=s[i];
          }
     }
     for(int i=;i<=n;i++){
         scanf("%lf",&p[i]);
     }
     int now;
     for(int i=;i<=;i++){//初始化
         now=(a[]>>(i-));
         if(now%==){
             dp[][i][]=1.0;
             dp[][i][]=0.0;
         }
         else{
             dp[][i][]=0.0;
             dp[][i][]=1.0;
         }
     }
     for(int i=;i<=n;i++){
         for(int j=;j<=;j++){
          dp[i][j][]+=dp[i-][j][]*p[i];//消失
          dp[i][j][]+=dp[i-][j][]*p[i];
         }
         if(o[i]=='^'){
             for(int j=;j<=;j++){//不消失
                 now=(a[i]>>(j-));
                 if(now%==){
                     dp[i][j][]+=dp[i-][j][]*(1.0-p[i]);
                     dp[i][j][]+=dp[i-][j][]*(1.0-p[i]);
                 }
                 else
                 {
                     dp[i][j][]+=dp[i-][j][]*(1.0-p[i]);
                     dp[i][j][]+=dp[i-][j][]*(1.0-p[i]);
                 }
             }
         }
          if(o[i]=='|'){
                 for(int j=;j<=;j++){
                 now=(a[i]>>(j-));
                 if(now%==){
                     dp[i][j][]+=(dp[i-][j][]+dp[i-][j][])*(1.0-p[i]);               }
                 else
                 {
                     dp[i][j][]+=dp[i-][j][]*(1.0-p[i]);
                     dp[i][j][]+=dp[i-][j][]*(1.0-p[i]);
                 }
             }

         }
          if(o[i]=='&'){
                 for(int j=;j<=;j++){
                 now=(a[i]>>(j-));
                 if(now%==){
                     dp[i][j][]+=dp[i-][j][]*(1.0-p[i]);
                     dp[i][j][]+=dp[i-][j][]*(1.0-p[i]);
                 }
                 else
                 {
                     dp[i][j][]+=(dp[i-][j][]+dp[i-][j][])*(1.0-p[i]);
                 }
             }
         }
     }
     double ans=;
     now=;
     for(int i=;i<=;i++){
         ans=ans+(dp[n][i][])*now;
         now*=;
     }
     printf("%.6f\n",ans);
     return ;
 }