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

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+10, M = 1e3+20,inf = 2e9;int n,a[N];
char c[N];
double dp[N][30][2],p[N];
int main() {while(scanf("%d",&n)!=EOF) {for(int i = 0; i <= n; ++i) {scanf("%d",&a[i]);}memset(dp,0,sizeof(dp));for(int i = 1; i <= n; ++i) {getchar();scanf("%c",&c[i]);}for(int i = 1; i <= n; ++i) {scanf("%lf",&p[i]);}for(int i = 0; i <= 21; ++i) {if(((1<<i)&a[0])) dp[0][i][1] = 1,dp[0][i][0] = 0;else dp[0][i][0] = 1,dp[0][i][1] = 0;}for(int i = 1; i <= n; ++i) {for(int j = 0; j <= 21; ++j) {dp[i][j][1] += 1.0*dp[i-1][j][1] * p[i];dp[i][j][0] += 1.0*dp[i-1][j][0] * p[i];}for(int j = 0; j <= 21; ++j) {int tmp = ((a[i]>>j)&1);if(c[i] == '^') {dp[i][j][tmp^1] += 1.0*dp[i-1][j][1]*(1.0-p[i]);dp[i][j][tmp^0] += 1.0*dp[i-1][j][0]*(1.0-p[i]);}else if(c[i] == '&'){dp[i][j][tmp&1] += 1.0*dp[i-1][j][1]*(1.0-p[i]);dp[i][j][tmp&0] += 1.0*dp[i-1][j][0]*(1.0-p[i]);}else if(c[i] == '|') {dp[i][j][tmp|1] += 1.0*dp[i-1][j][1]*(1.0-p[i]);dp[i][j][tmp|0] += 1.0*dp[i-1][j][0]*(1.0-p[i]);}}}double ans = 0;for(int i = 0; i <= 21; ++i) {LL tmp = 1<<i;ans += (double)(dp[n][i][1]) * 1.0 * tmp;}printf("%.6f\n",ans);}return 0;
}

转载于:https://www.cnblogs.com/zxhl/p/7256376.html

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