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:
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 probability0.5).
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?
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 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.
++-+- +-+-+
1.000000000000
+-+- +-??
0.500000000000
+++ ??-
0.000000000000
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <iostream>
using namespace std;
char buf1[15];
char buf2[15];
int aim;
int cnt;
void dfs(int left, int sum)
{
if (left == 0)
{
if (sum == aim)
cnt++;
return;
}
if (aim - sum <= left && aim - sum >= -left)
{
dfs(left-1, sum+1);
dfs(left-1, sum-1);
}
}
int main()
{
cin >> buf1 >> buf2;
int unrec = 0, v1 = 0, v2 = 0;
int len = strlen(buf1);
for(int i=0; i<=len-1; i++)
{
if (buf1[i] == '+')
++v1;
else
--v1;
}
for(int i=0; i<=len-1; i++)
{
if (buf2[i] == '+')
++v2;
else if (buf2[i] == '-')
--v2;
else
++unrec;
}
aim = v1 - v2;
cnt = 0;
dfs(unrec, 0);
int tot = 1;
for(int i=1;i<=unrec;i++)
tot *= 2;
printf("%.12f", (double)cnt/(double)tot);
return 0;
}Codeforces Round #272 (Div. 2) B
原文：http://blog.csdn.net/notdeep__acm/article/details/40051625