hdu 4649 Professor Tian(期望)

Professor Tian

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 932    Accepted Submission(s): 523

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.

 

Sample Input

21 2 3^ ^0.1 0.228 9 10^ ^0.5 0.7811 2&0.5

 

Sample Output

Case 1:0.720000Case 2:4.940000Case 3:0.500000

题意：给出一个二进制点的位运算，和每个位运算数字和符号出现的概率，求答案的期望；

解：本质上就是求一个二进制数的每一位的是1的概率对答案的贡献，为0的不考虑

#include<iostream>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<vector>#include<map>#include <bits/stdc++.h>using namespace std;const int N = 1100000+10;typedef long long LL;const LL mod = 1e9+7;char str[N][3];int a[N];double p[N];int main(){    int ncase=1, n;    while(scanf("%d", &n)!=EOF)    {        for(int i=0;i<=n;i++) scanf("%d", &a[i]);        for(int i=1;i<=n;i++) scanf("%s",str[i]);        for(int i=1;i<=n;i++) scanf("%lf", &p[i]);        double ans=0, q;        for(int i=0;i<=20;i++)        {            if(a[0]&(1<<i)) q=1;            else q=0;            for(int j=1;j<=n;j++)            {                if(a[j]&(1<<i))                {                    if(str[j][0]=='^') q=q*p[j]+(1-q)*(1-p[j]);                    else if(str[j][0]=='|') q=q*p[j]+(1-p[j]);                }                else                {                    if(str[j][0]=='&') q=q*p[j];                }            }            ans+=(1<<i)*q;        }        printf("Case %d:\n",ncase++);        printf("%.6f\n",ans);    }    return 0;}