Codeforces round #272 (Div. 2) B

Question: B. dreamoon and wifitime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard 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 ction, denoted as '+'
2. Go 1 unit towards the negative ction, denoted '-'

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 stored ed 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 stringS1-The commands drazil sends to dreamoon, this string consists of only the characters in the set {'+ ','-'}.

The second line contains a stringS2-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

Meaning analysis: a string. "+" indicates a step forward, and "-" indicates a step forward. Returns the second string. The addition and subtraction numbers have the same meaning. There are multiple "?" strings. Indicates that both the front and back are allowed. Calculate the probability that the positions are the same based on the two solutions. The first string can directly simulate the final position, and then the second string needs to be searched. The probability is equal to the first case except all cases.
Code:

#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;}

Codeforces round #272 (Div. 2) B