Codeforces Round #272 (Div. 2) B

题目：

B. Dreamoon and WiFi

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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:

1. Go 1 unit towards the positive direction, denoted as '+'
2. 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 probability0.5).

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 test(s)

input

++-+-
+-+-+

output

1.000000000000

input

+-+-
+-??

output

0.500000000000

input

+++
??-

output

0.000000000000

题意分析：

给出一个字符串，“+”表示往前走一步，“-”表示往后走一步。给出第二个字符串，加减号含义一样，多个“？”表示往前后都可以。求最后按照两种方案走，位置相同的概率。

第一个字符串直接能模拟出最后的位置，然后第二个需要搜索一下，等于第一种的情况除所有情况数就是概率。

代码：

#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <iostream>

using namespace std;

char buf1[15];
char buf2[15];
int aim;
int cnt;

void dfs(int left, int sum)
{
    if (left == 0)
    {
        if (sum == aim)
            cnt++;
        return;
    }
    if (aim - sum <= left && aim - sum >= -left)
    {
        dfs(left-1, sum+1);
        dfs(left-1, sum-1);
    }
}

int main()
{

    cin >> buf1 >> buf2;
    int unrec = 0, v1 = 0, v2 = 0;
    int len = strlen(buf1);
    for(int i=0; i<=len-1; i++)
    {
        if (buf1[i] == '+')
            ++v1;
        else
            --v1;
    }
    for(int i=0; i<=len-1; i++)
    {
        if (buf2[i] == '+')
            ++v2;
        else if (buf2[i] == '-')
            --v2;
        else
            ++unrec;
    }
    aim = v1 - v2;
    cnt = 0;
    dfs(unrec, 0);
    int tot = 1;
    for(int i=1;i<=unrec;i++)
        tot *= 2;
    printf("%.12f", (double)cnt/(double)tot);
    return 0;
}