Codeforces Round 272 (Div. 2)
B
Dreamoon and WiFi
Codeforces
1300
原题
← AC →
翻译中…

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.

翻译中…

Each command is one of the following two types:

翻译中…
  1. Go 1 unit towards the positive direction, denoted as '+'
  2. Go 1 unit towards the negative direction, denoted as '-'
翻译中…

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 $0.5$).

翻译中…

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?

翻译中…

Input

翻译中…

The first line contains a string $*s*_{1}$ — the commands Drazil sends to Dreamoon, this string consists of only the characters in the set {'+', '-'}.

翻译中…

The second line contains a string $*s*_{2}$ — the commands Dreamoon's smartphone recognizes, this string consists of only the characters in the set {'+', '-', '?'}. '?' denotes an unrecognized command.

翻译中…

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

翻译中…

Output

翻译中…

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

翻译中…

Note

翻译中…

For the first sample, both $*s*_{1}$ and $*s*_{2}$ will lead Dreamoon to finish at the same position $+ 1$.

翻译中…

For the second sample, $*s*_{1}$ will lead Dreamoon to finish at position 0, while there are four possibilites for $*s*_{2}$: {"+-++", "+-+-", "+--+", "+---"} 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 $0.5$.

翻译中…

For the third sample, $*s*_{2}$ could only lead us to finish at positions {+1, -1, -3}, so the probability to finish at the correct position $+ 3$ is $0$.

翻译中…

样例

Input #1
++-+-
+-+-+
Output #1
1.000000000000
InputCopy OutputCopy
Input #2
+-+-
+-??
Output #2
0.500000000000
InputCopy OutputCopy
Input #3
+++
??-
Output #3
0.000000000000
InputCopy OutputCopy
题目信息
题号
476B
时间限制
1 s
内存限制
256 MB
标签
位运算
暴力
组合数学
动态规划
数学
概率论

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