codeforces Round #272(div2) B解题报告

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 for s2: {"+-++", "+-+-", "+--+", "+---"} 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 is0.5.

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：

给出Drazil的命令序列和Dreamoon获得的序列，求出Dreamoon到达指定位置的概率。

解法：

先根据Drazil的命令序列处理出指定的位置“rigpos”，然后假设Dreamoon先不管“?”命令，先执行准确的命令，求出Dreamoon根据准确命令可以到达的位置"wepos"。（命令执行的先后不影响最终的位置）。

然后，从wepos开始概率为1，然后用bfs从左右扩散概率（代码没用bfs，因为数据很小），最终可以求出到达指定位置的概率。

代码：

#include <cstdio>#include <cstring>#define esp 1e-10using namespace std;double ans[100][100];char st1[20], st2[20];int rigpos, wepos, num;void init() {scanf("%s", st1);scanf("%s", st2);for (int i = 0; i < strlen(st1); i++)if (st1[i] == '+')rigpos++;elserigpos--;for (int i = 0; i < strlen(st2); i++)if (st2[i] == '-')wepos--;else if (st2[i] == '+')wepos++;elsenum++;}void solve() {ans[0][wepos+50] = 1;for (int i = 0; i < num; i++)for (int j = 30; j < 80; j++) {ans[i+1][j-1] += ans[i][j]*0.5;ans[i+1][j+1] += ans[i][j]*0.5;}printf("%.12lf", ans[num][rigpos+50]);}int main() {init();solve();}