B. Dreamoon and WiFi

题目连接:

http://www.codeforces.com/contest/476/problem/B

Description

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:

Go 1 unit towards the positive direction, denoted as '+'

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 Input

++-+-

+-+-+

Sample Output

1.000000000000

Hint

题意

给你一个串s1,和一个串s2

然后s2中有一些问号,有0.5的概率是向左边,有0.5的概率是向右边走。

问你s2结束之后,恰好走到s1串走到的目的地的概率是多少。

题解:

数据范围很小,怎么做都行。

我是概率dp,dp[i][j]表示考虑到执行第二个串的第i位现在在j位置的概率是多少。

代码

#include<bits/stdc++.h>
using namespace std; double dp[100][100];
char s1[50],s2[50];
int main()
{
scanf("%s%s",s1+1,s2+1);
int level = 50;
int pos=50;
int len1 = strlen(s1+1);
for(int i=1;i<=len1;i++)
{
if(s1[i]=='+')pos++;
else pos--;
}
dp[0][level]=1;
for(int i=1;i<=len1;i++)
{
if(s2[i]=='+')for(int j=1;j<100;j++)
dp[i][j]=dp[i-1][j-1];
if(s2[i]=='-')for(int j=0;j<99;j++)
dp[i][j]=dp[i-1][j+1];
if(s2[i]=='?')for(int j=1;j<99;j++)
dp[i][j]=0.5*dp[i-1][j-1]+0.5*dp[i-1][j+1];
}
printf("%.12f\n",dp[len1][pos]);
}

Codeforces Round #272 (Div. 2) B. Dreamoon and WiFi dp的更多相关文章

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

    http://codeforces.com/contest/476/problem/B B. Dreamoon and WiFi time limit per test 1 second memory ...

  2. Codeforces Round #272 (Div. 2) E. Dreamoon and Strings dp

    题目链接: http://www.codeforces.com/contest/476/problem/E E. Dreamoon and Strings time limit per test 1 ...

  3. Codeforces Round #272 (Div. 2) B. Dreamoon and WiFi (暴力二进制枚举)

    题意:给你一个只含\(+\)和\(-\)的字符串,统计它的加减和,然后再给你一个包含\(+,-,?\)的字符串,其中\(?\)可以表示为\(+\)或\(-\),问有多少种情况使得第二个字符串的加减和等 ...

  4. Codeforces Round #272 (Div. 2) E. Dreamoon and Strings 动态规划

    E. Dreamoon and Strings 题目连接: http://www.codeforces.com/contest/476/problem/E Description Dreamoon h ...

  5. Codeforces Round #272 (Div. 2) D. Dreamoon and Sets 构造

    D. Dreamoon and Sets 题目连接: http://www.codeforces.com/contest/476/problem/D Description Dreamoon like ...

  6. Codeforces Round #272 (Div. 2) A. Dreamoon and Stairs 水题

    A. Dreamoon and Stairs 题目连接: http://www.codeforces.com/contest/476/problem/A Description Dreamoon wa ...

  7. Codeforces Round #272 (Div. 1) A. Dreamoon and Sums(数论)

    题目链接 Dreamoon loves summing up something for no reason. One day he obtains two integers a and b occa ...

  8. Codeforces Round #272 (Div. 2)-C. Dreamoon and Sums

    http://codeforces.com/contest/476/problem/C C. Dreamoon and Sums time limit per test 1.5 seconds mem ...

  9. Codeforces Round #272 (Div. 2)-A. Dreamoon and Stairs

    http://codeforces.com/contest/476/problem/A A. Dreamoon and Stairs time limit per test 1 second memo ...

随机推荐

  1. hdu 5181 numbers

    http://acm.hdu.edu.cn/showproblem.php?pid=5181 题意: 有一个栈,其中有n个数1~n按顺序依次进入栈顶,在某个时刻弹出. 其中m个限制,形如数字A必须在数 ...

  2. [iOS]Xcode+GitHub远程代码托管(GIT, SVN)

    先来看看什么是代码远程托管: 其实就是将我们的代码上传到GitHub的服务器上, 供别人下载, 当然了也可以在团队开发的时候, 使用GitHub进行代码合并工作, 下面我们进入正题 (已经有远程仓库的 ...

  3. 兼容 IE6+ 获取图片大小

    昨天说了 HTML5 的file对象可以获取到文件的 文件名,文件大小,文件类型,最后一次修改日期.其实 IE6-9 也可以变向的获取到,虽然没那么方便,但至少可以取到.来看例子吧: <!doc ...

  4. javascript柯里化

    function curry(fn){ var slice = Array.prototype.slice; var arr = slice.call(arguments,1); return fun ...

  5. ASP.net 控件实现数据级联

    今天我们来一起用ASP.net实现一个级联,这个小不点应该是会经常用到的的. 咱们简单的画两个窗体.文本框会根据下拉框所选的内容显示不同的内容. 具体实现效果如下 步骤一: 准备工作,建立相应的数据库 ...

  6. Linux - awk 文本处理工具二

    awk 判断格式 awk '{print ($1>$2)?"第一排"$1:"第二排"$2}' # 条件判断 括号代表if语句判断 "?" ...

  7. linux - JDK 环境

    JDK安装 vi /etc/profile # 添加环境变量 export JAVA_HOME=/usr/local/jdk1.8.2_45 export CLASSPATH=.:$JAVA_HOME ...

  8. 【洛谷P3884 [JLOI2009]】二叉树问题

    题目描述 如下图所示的一棵二叉树的深度.宽度及结点间距离分别为: 深度:4 宽度:4(同一层最多结点个数) 结点间距离: ⑧→⑥为8 (3×2+2=8) ⑥→⑦为3 (1×2+1=3) 注:结点间距离 ...

  9. Route Between Two Nodes in Graph

    Given a directed graph, design an algorithm to find out whether there is a route between two nodes. ...

  10. 转:vue2.0 keep-alive最佳实践

    转载至:https://segmentfault.com/a/1190000008123035 1.基本用法 vue2.0提供了一个keep-alive组件用来缓存组件,避免多次加载相应的组件,减少性 ...