lonlifeOJ1152 “玲珑杯”ACM比赛 Round #19 概率DP

E -- Expected value of the expression

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

 

 

题意：

　　给你一个n+1个数进行位操作

　　给你这个n+1个数(a0~an)和 进行的操作(异或，并，或) c[i]

　　ci 和 ai同时消失的 概率是 pi

　　求最后值得期望

题解：

　　dp[i][25][0/1]

　　表示前i个 数  0~21位上每一位存在（0/1）的概率，强推过去就行了

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18+1LL;
const double pi = acos(-1.0);
const int N = 5e3+, M = 1e3+,inf = 2e9;

int n,a[N];
char c[N];
double dp[N][][],p[N];
int main() {
    while(scanf("%d",&n)!=EOF) {
        for(int i = ; i <= n; ++i) {
            scanf("%d",&a[i]);
        }
        memset(dp,,sizeof(dp));
        for(int i = ; i <= n; ++i) {
            getchar();
            scanf("%c",&c[i]);
        }
        for(int i = ; i <= n; ++i) {
            scanf("%lf",&p[i]);
        }
        for(int i = ; i <= ; ++i) {
            if(((<<i)&a[])) dp[][i][] = ,dp[][i][] = ;
            else dp[][i][] = ,dp[][i][] = ;
        }
        for(int i = ; i <= n; ++i) {

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