Codeforces Round #272 (Div. 2)B. Dreamoon and WiFi

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 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 test(s)

input

++-+-+-+-+

output

1.000000000000

input

+-+-+-??

output

0.500000000000

input

+++??-

output

0.000000000000

Note

For the first sample, both s1 and s2 will lead Dreamoon to finish at the same position  + 1.

For the second sample, s1 will lead Dreamoon to finish at position 0, while there are four possibilites fors2: {"+-++", "+-+-", "+--+", "+---"} with ending position {+2, 0, 0, -2} respectively. So there are 2 correct cases out of 4, so the probability of finishing at the correct position is 0.5.

For the third sample, s2 could only lead us to finish at positions {+1, -1, -3}, so the probability to finish at the correct position  + 3 is 0.

题目大意：一道有关搜索的题目，说的是一个人从0点开始，在一条线段上移动（嘿嘿，好久不做这样的一维线段问题了），然后，"+"表示向右移动一个位置，而“-”表示向左边移动一个位置，“？”表示两种位置移动的可能行都有，为最终在所有的移动操作都结束后，他回到原点的概率是多少，

也就是说第一个串的操作是正确的，第一个串操作结束后，会产生一个具体的坐标位置，这个坐标位置就是我们要用第二个串作为检验的结果，由于第二个串的输入有三种情况，也就是说，可能会产生“+”，“-”，“？”，故这样的变化就带来了随机事件的产生，所以自然而然引出了概率，概率其实很简单，不用虚，简单的古典概型，高中筒子都能做的题。

输出看的有些2b~，注意不要PE就行了。

算法思想：将所有可能的路径DFS一遍，然后暴力计数就好~

代码如下：

# include<cstdio># include<iostream># include<cstring>using namespace std;char op[15];int ans1,ans2;int sum[30];void dfs( int pos,int count ){    if ( pos == strlen(op) )    {        sum[count+10]++;        ans2++;        return;    }    if ( op[pos] == '+' )    {        dfs(pos+1,count+1);    }    else if ( op[pos] == '-' )    {        dfs(pos+1,count-1);    }    else    {        dfs(pos+1,count+1);        dfs(pos+1,count-1);    }}int main(void){    while( cin>>op )    {        ans1 = 0;        ans2 = 0;        int len = strlen(op);         for ( int i = 0;i < len;i++ )         {             if ( op[i] == '+' )                ans1++;                else                ans1--;            }            cin>>op;            dfs(0,0);            printf("%.10f\n",sum[ans1+10]/ans2*1.0);    }    return 0;}