题解

#include <iostream>
#include <string>
#include <iomanip>

using namespace std;

// 计算组合数 C(n, k)
long long comb(int n, int k) {
    if (k < 0 || k > n) return 0;
    if (k == 0 || k == n) return 1;
    k = min(k, n - k); // 利用对称性减少计算量
    long long res = 1;
    for (int i = 1; i <= k; ++i) {
        res = res * (n - k + i) / i;
    }
    return res;
}

int main() {
    string original, received;
    cin >> original >> received;

    // 计算原始指令的目标位置
    int target = 0;
    for (char c : original) {
        target += (c == '+') ? 1 : -1;
    }

    // 计算收到指令的确定位置和问号数量
    int pos = 0;
    int q_cnt = 0;
    for (char c : received) {
        if (c == '+') pos++;
        else if (c == '-') pos--;
        else q_cnt++;
    }

    // 计算需要的净位移
    int delta = target - pos;

    // 计算概率
    double prob = 0.0;
    if (abs(delta) <= q_cnt && (q_cnt + delta) % 2 == 0) {
        int m = (q_cnt + delta) / 2;
        long long c = comb(q_cnt, m);
        prob = static_cast<double>(c) / (1LL << q_cnt);
    }

    // 输出结果，保留6位小数
    cout << fixed << setprecision(6) << prob << endl;

    return 0;
}