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: ${\mathrm{A0O1A1O2A2\cdots OnAnA0O1A1O2A2\cdots OnAn,\; where\; Ai(0\le i\le n)Ai(0\le i\le n)\; represents\; number,\; Oi(1\le i\le n)Oi(1\le i\le n)\; represents\; operator.\; There\; are\; three\; operators,\; \&,|,^\&,|,^,\; which\; means\; and,or,xorand,or,xor,\; and\; they\; have\; the\; same\; priority.}}^{}$

The $\mathrm{ii-th\; operator\; OiOi\; and\; the\; numbers\; AiAi\; disappear\; with\; the\; probability\; of\; pipi.}$

Find the expected value of an expression.

INPUT

The first line contains only one integer $\mathrm{n(1\le n\le 1000)n(1\le n\le 1000).\; The\; second\; line\; contains\; n+1n+1\; integers\; Ai(0\le Ai<220)Ai(0\le Ai<220).\; The\; third\; line\; contains\; nn\; chars\; OiOi.\; The\; fourth\; line\; contains\; nn\; floats\; pi(0\le pi\le 1)pi(0\le pi\le 1).}$

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[2000000];
int a[2000];
char o[2000];
double p[2000];
double dp[2000][22][2];
int main()
{
    scanf("%d",&n);
    for(int i=0;i<=n;i++){
        scanf("%d",&a[i]);
    }
    getchar();
    gets(s);
    int len=strlen(s);
    int res=1;
    for(int i=0;i<len;i++){
        if(s[i]!=' '){
         o[res++]=s[i];
         }
    }
    for(int i=1;i<=n;i++){
        scanf("%lf",&p[i]);
    }
    int now;
    for(int i=1;i<=21;i++){//初始化
        now=(a[0]>>(i-1));
        if(now%2==1){
            dp[0][i][1]=1.0;
            dp[0][i][0]=0.0;
        }
        else{
            dp[0][i][1]=0.0;
            dp[0][i][0]=1.0;
        }
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=21;j++){
         dp[i][j][0]+=dp[i-1][j][0]*p[i];//消失
         dp[i][j][1]+=dp[i-1][j][1]*p[i];
        }
        if(o[i]=='^'){
            for(int j=1;j<=21;j++){//不消失
                now=(a[i]>>(j-1));
                if(now%2==1){
                    dp[i][j][1]+=dp[i-1][j][0]*(1.0-p[i]);
                    dp[i][j][0]+=dp[i-1][j][1]*(1.0-p[i]);
                }
                else
                {
                    dp[i][j][1]+=dp[i-1][j][1]*(1.0-p[i]);
                    dp[i][j][0]+=dp[i-1][j][0]*(1.0-p[i]);
                }
            }
        }
         if(o[i]=='|'){
                for(int j=1;j<=21;j++){
                now=(a[i]>>(j-1));
                if(now%2==1){
                    dp[i][j][1]+=(dp[i-1][j][0]+dp[i-1][j][1])*(1.0-p[i]);               }
                else
                {
                    dp[i][j][1]+=dp[i-1][j][1]*(1.0-p[i]);
                    dp[i][j][0]+=dp[i-1][j][0]*(1.0-p[i]);
                }
            }

        }
         if(o[i]=='&'){
                for(int j=1;j<=21;j++){
                now=(a[i]>>(j-1));
                if(now%2==1){
                    dp[i][j][1]+=dp[i-1][j][1]*(1.0-p[i]);
                    dp[i][j][0]+=dp[i-1][j][0]*(1.0-p[i]);
                }
                else
                {
                    dp[i][j][0]+=(dp[i-1][j][0]+dp[i-1][j][1])*(1.0-p[i]);
                }
            }
        }
    }
    double ans=0;
    now=1;
    for(int i=1;i<=21;i++){
        ans=ans+(dp[n][i][1])*now;
        now*=2;
    }
    printf("%.6f\n",ans);
    return 0;
}