codeforces B. Dreamoon and WiFi 組合數學求概率

codeforces B. Dreamoon and WiFi 組合數學求概率

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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.

   
   給定兩個字符串，第一個字符串由"+","-"組成，第二個字符串由"+","-","?"組成，“+”代表加1，"-"代表減一，“?"代表可取正也可取負，問第二個字符串和第一個字符串相等的概率是多少。 設第二個字符串"+"的個數為x，"-"的個數為y，"?"的個數為c,第一個字符串值為a，第二個字符串值為b。可以聯立方程組：x+y=c① x-y=fabs（a-b)② 求出x，若x無法求得，則概率為0。再根據排列組合公式求出最終概率C(c,x)*pow(0.5,x)*pow(0.5,c-y)。

   #include
   #include
   #include
   char s1[17],s2[17];
   int C(int c,int x)
   {
       int a=1,b=1,d=1;
       for(int i=1;i<=c;i++)a*=i;
       for(int i=1;i<=x;i++)b*=i;
       for(int i=1;i<=(c-x);i++)d*=i;
       return a/(b*d);
   }
   int main()
   {
       while(scanf("%s%s",s1,s2)!=EOF)
       {
           int len1=strlen(s1);
           int len2=strlen(s2);
           int a=0,b=0,c=0;
           for(int i=0;i>1;
                   int aa=C(c,x);
                   printf("%.12lf\n",aa*(double)(pow(0.5,x))*(double)(pow(0.5,(c-x))));
               }
           }
       }
       return 0;
   }