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

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 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.

解题思路: 求接收到的信号与初始信号直接相同的概率

所以只要记录下接收到的有几个不确定的(即‘?’的个数),然后依次枚举即可

#include <stdio.h>
#include <string.h>
#include <stdlib.h> int main(){
    char str1[], str2[];
    int num1[], num2[], len, sum, i, t, x, y, n, m, flag;
    while(scanf("%s %s", str1, str2) != EOF){
        len =strlen(str1);
        sum = ;
        for(i = ; i < len; i++){
            num1[i] = str1[i] == '+' ?  : ;
            if(str2[i] == '+'){
                num2[i] = ;
            }
            else if(str2[i] == '-'){
                num2[i] = ;
            }
            else if(str2[i] == '?'){
                sum++;
            }
        }
        x = ;
        flag = ;
        for(i = ; i < len; i++){
            x += num1[i] ==  ?  : - ;
        }
        for(i = ; i < sum; i++){
            flag *= ;
        }
        t = flag;
        n = m = ;
        while(t > ){
            for(i = ; i < len; i++){
                if(str2[i] == '?'){
                    if(t %  == ){
                        num2[i] = ;
                    }
                    else{
                        num2[i] = ;
                    }
                    t /= ;
                }
            }
            y = ;
            for(i = ; i < len; i++){
                y += (num2[i] ==  ?  : -);
            }
            t = --flag;
            if(x == y){
                n++;
            }
            m++;
        }
        printf("%.12lf\n",(double)n / (double)m);
    }
    return ;
}

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

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

    B. Dreamoon and WiFi 题目连接: http://www.codeforces.com/contest/476/problem/B Description Dreamoon is s ...

  2. 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 ...

  3. 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 ...

  4. 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 ...

  5. 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 ...

  6. 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 ...

  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. 7 二分搜索树的原理与Java源码实现

    1 折半查找法 了解二叉查找树之前,先来看看折半查找法,也叫二分查找法 在一个有序的整数数组中(假如是从小到大排序的),如果查找某个元素,返回元素的索引. 如下: int[] arr = new in ...

  2. Weekly Contest 78-------->808. Soup Servings

    There are two types of soup: type A and type B. Initially we have N ml of each type of soup. There a ...

  3. lightoj 1027【数学概率】

    #include <bits/stdc++.h> using namespace std; typedef long long LL; const int N=1e2+10; int ma ...

  4. CodeForces Canada Cup 2016【A,B,C,D】

    CodeForces 725A: 思路就是如果"最左"不是'>'这个了,那么这个右边的一定不可能到达左边了: 同理最右: CodeForces 725B: 有两个空姐,一个从 ...

  5. A. Office Keys ( Codeforces Round #424 (Div. 1, rated, based on VK Cup Finals) )

    #include <iostream> #include <stdio.h> #include <string.h> #include <algorithm& ...

  6. with rollup

    实验吧的一道ctf题,这两天无聊,做做ctf题.在实验吧被一道也题卡了好久. 页面很简单就是一个登陆页面,按照之前的经验觉得应该是注入吧.再看题猜测应该是绕waf之类的. 查看页面源码找到了提供的源代 ...

  7. 牛客寒假6-C.项链

    链接:https://ac.nowcoder.com/acm/contest/332/C 题意: 小B想给她的新项链染色. 现在有m种颜色,对于第i种颜色,小B有a_i单位的颜料,每单位颜料可以染项链 ...

  8. 【aspnetcore】让aspnetcore支持less文件

    第一步:新建文件 CustomerFileExtensionContentTypeProvider namespace xxx { public class CustomerFileExtension ...

  9. Springboot日志配置探索(主要看logback)(一)

    这篇博客是springboot日志配置探索的第一篇,主要讲默认配置下springboot的logback日志框架的配置(即直接使用是怎样的) 首先,是一个SpringBoot的有关日志的说明文档:ht ...

  10. python转换已转义的字符串

    python转换已转义的字符串 有时我们可能会获取得以下这样的字符串: >>> a = '{\\"name\\":\\"michael\\"} ...