“玲珑杯”ACM比赛 Round #19 E.Expected value of the expression【Dp】

1152 - Expected value of the expression

Time Limit：2s Memory Limit：128MByte

Submissions：82Solved：45

DESCRIPTION

You are given an expression: $A_0{O}_1{A}_1{O}_2{A}_2\cdots O_n{A}_n$, where $A_i(0\le i\le n)$ represents number, $O_i(1\le i\le n)$ represents operator. There are three operators, $\&,|,\hat{}$, which means $\text{and},\text{or},\text{xor}$, and they have the same priority.

The $i$-th operator $O_i$ and the numbers $A_i$ disappear with the probability of $p_i$.

Find the expected value of an expression.

INPUT

The first line contains only one integer$n(1\le n\le 1000)$.The second line contains $n+1$ integers $A_i(0\le A_i<2^{20})$.The third line contains $n$ chars $O_i$.The fourth line contains $n$ floats $p_i(0\le p_i\le 1)$.

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

题目大意：

有N+1个数Ai.有N个位运算操作，在其中间隔排列，当前位运算操作出现的概率是1-pi，问最终获得的值的期望。

思路：

按位dp.分成20位、设定Dp【i】【2】:

①Dp【i】【0】表示对应当前位，到了第i个操作符，运算结果为0的概率

②Dp【i】【1】表示对应当前位，到了第i个操作符，运算结果为1的概率

那么状态转移方程我们到当前位i分两种情况考虑即可：

①当前操作符不粗线了。

②当前操作符出现。

然后再分操作符的三种情况转移一下就行了。

Ac代码：

#include<stdio.h>#include<string.h>using namespace std;int a[15000];int num[15000];char op[15000][3];double p[15000];double dp[15000][3];int main(){    int n;    while(~scanf("%d",&n))    {        for(int i=0;i<=n;i++)scanf("%d",&num[i]);        for(int i=1;i<=n;i++)scanf("%s",op[i]);        for(int i=1;i<=n;i++)scanf("%lf",&p[i]);        double output=0;        for(int i=0;i<=20;i++)        {            for(int j=0;j<=n;j++)            {                if((num[j]&(1<<i))>0)                {                    a[j]=1;                }                else a[j]=0;            }            memset(dp,0,sizeof(dp));            dp[0][a[0]]=1;            for(int j=1;j<=n;j++)            {                dp[j][0]=dp[j-1][0]*p[j];                dp[j][1]=dp[j-1][1]*p[j];                if(op[j][0]=='|')                {                    if(a[j]==0)                    {                        dp[j][0]+=dp[j-1][0]*(1-p[j]);                        dp[j][1]+=dp[j-1][1]*(1-p[j]);                    }                    if(a[j]==1)                    {                        dp[j][1]+=dp[j-1][0]*(1-p[j]);                        dp[j][1]+=dp[j-1][1]*(1-p[j]);                    }                }                if(op[j][0]=='&')                {                    if(a[j]==0)                    {                        dp[j][0]+=dp[j-1][0]*(1-p[j]);                        dp[j][0]+=dp[j-1][1]*(1-p[j]);                    }                    if(a[j]==1)                    {                        dp[j][0]+=dp[j-1][0]*(1-p[j]);                        dp[j][1]+=dp[j-1][1]*(1-p[j]);                    }                }                if(op[j][0]=='^')                {                    if(a[j]==0)                    {                        dp[j][0]+=dp[j-1][0]*(1-p[j]);                        dp[j][1]+=dp[j-1][1]*(1-p[j]);                    }                    if(a[j]==1)                    {                        dp[j][0]+=dp[j-1][1]*(1-p[j]);                        dp[j][1]+=dp[j-1][0]*(1-p[j]);                    }                }            }            output+=(dp[n][1]*(1<<i));        }        printf("%.6f\n",output);    }}