B. Dreamoon and WiFi

题目连接：

http://www.codeforces.com/contest/476/problem/B

Description

Dreamoon is standing at the position 0 on a number line. Drazil is sending a list of commands through Wi-Fi to Dreamoon's smartphone and Dreamoon follows them.

Each command is one of the following two types:

Go 1 unit towards the positive direction, denoted as '+'

Go 1 unit towards the negative direction, denoted as '-'

But the Wi-Fi condition is so poor that Dreamoon's smartphone reports some of the commands can't be recognized and Dreamoon knows that some of them might even be wrong though successfully recognized. Dreamoon decides to follow every recognized command and toss a fair coin to decide those unrecognized ones (that means, he moves to the 1 unit to the negative or positive direction with the same probability 0.5).

You are given an original list of commands sent by Drazil and list received by Dreamoon. What is the probability that Dreamoon ends in the position originally supposed to be final by Drazil's commands?

Input

The first line contains a string s1 — the commands Drazil sends to Dreamoon, this string consists of only the characters in the set {'+', '-'}.

The second line contains a string s2 — the commands Dreamoon's smartphone recognizes, this string consists of only the characters in the set {'+', '-', '?'}. '?' denotes an unrecognized command.

Lengths of two strings are equal and do not exceed 10.

Output

Output a single real number corresponding to the probability. The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 9.

Sample Input

++-+-

+-+-+

Sample Output

1.000000000000

Hint

题意

给你一个串s1，和一个串s2

然后s2中有一些问号，有0.5的概率是向左边，有0.5的概率是向右边走。

问你s2结束之后，恰好走到s1串走到的目的地的概率是多少。

题解：

数据范围很小，怎么做都行。

我是概率dp,dp[i][j]表示考虑到执行第二个串的第i位现在在j位置的概率是多少。

代码

#include<bits/stdc++.h>

using namespace std;

double dp[100][100];

char s1[50],s2[50];

int main()

{

    scanf("%s%s",s1+1,s2+1);

    int level = 50;

    int pos=50;

    int len1 = strlen(s1+1);

    for(int i=1;i<=len1;i++)

    {

        if(s1[i]=='+')pos++;

        else pos--;

    }

    dp[0][level]=1;

    for(int i=1;i<=len1;i++)

    {

        if(s2[i]=='+')for(int j=1;j<100;j++)

            dp[i][j]=dp[i-1][j-1];

        if(s2[i]=='-')for(int j=0;j<99;j++)

            dp[i][j]=dp[i-1][j+1];

        if(s2[i]=='?')for(int j=1;j<99;j++)

            dp[i][j]=0.5*dp[i-1][j-1]+0.5*dp[i-1][j+1];

    }

    printf("%.12f\n",dp[len1][pos]);

}