Codeforces Gym 101158 E. Infallibly Crack Perplexing Cryptarithm （模拟 + 语法分析）

AC代码

C++版本

/*
根据 定义的等式 的规则
判断给定串可以写成多少种二进制等式且成立的形式

解法：
通过语法分析的形式将多个表达式定义进行 递归下降的解析。
此方法可以不用过多考虑多种表达式定义组合的合法和非法情况，只需要完整解析单个表达式，并递归调用。
*/

#include
using namespace std;
#define Result pair
#define FAIL make_pair((char*)NULL,0)
char srcExpr[32], expr[32], op[8] = {'0', '1', '+', '-', '*', '(', ')', '='};
struct Parser {
    Result Q(char* p) {
        Result res = E(p);
        if(res.first == NULL || *(res.first) != '=')    return FAIL;

        Result rgt = E(res.first+1);
        if(rgt.first == NULL || *(rgt.first) != 0 || rgt.second != res.second)  return FAIL;
        return rgt;
    }
    Result E(char *p) {
        Result ret = T(p);
        if(ret.first == NULL)   return FAIL;
        while(*(ret.first) == '+' || *(ret.first) == '-')
        {
            Result tmp = T(ret.first + 1);
            if(tmp.first == NULL)   return FAIL;
            if(*(ret.first) == '-') ret.second -= tmp.second;
            else    ret.second += tmp.second;
            ret.first = tmp.first;
        }
        return ret;
    }
    Result T(char *p) {
        Result ret = F(p);
        if(ret.first == NULL)   return FAIL;
        if(*(ret.first) == '*')
        {
            Result tmp = T(ret.first + 1);
            if(tmp.first == NULL)   return FAIL;
            ret.first = tmp.first,  ret.second *= tmp.second;
        }
        return ret;
    }
    Result F(char *p) {
        Result ret;
        if(*p == '-') {
            Result ret = F(p+1);
            ret.second = -ret.second;
            return ret;
        } else if(*p == '(') {
            Result ret = E(p+1);
            if(ret.first == NULL || *(ret.first) != ')')    return FAIL;
            ret.first++;
            return ret;
        } else {
            return N(p);
        }
    }
    Result N(char *p) {
        Result ret;
        if(!isdigit(*p))    return FAIL;
        if(*p == '0' && isdigit(*(p+1)))    return FAIL;
        while(isdigit(*p))
        {
            (ret.second *= 2) += (*p-'0');
            p++;
        }
        ret.first = p;
        return ret;
    }
};
int main()
{
    sort(op, op+8);
    scanf("%s", srcExpr);
    map<char, int> mp;
    int idx = 0;
    for(int i=0;srcExpr[i];i++)
    {
        if(isalpha(srcExpr[i]) && mp.find( srcExpr[i] ) == mp.end())
            mp[ srcExpr[i] ] = ++idx;
    }

    if(idx > 8) {   printf("0\n");  return 0;   }
    int ans = 0;
    do {
        for(int i=0;srcExpr[i];i++)
            expr[i] = (isalpha(srcExpr[i]) ? op[ mp[srcExpr[i]]-1 ] : srcExpr[i]);
        if(Parser().Q(expr).first != NULL)  ans++;  
    } while(next_permutation(op, op+8));
    int factorial = 1;
    for(int i=1;i<=(8-idx);i++)
        factorial *= i;
    printf("%d\n", ans / factorial);
}

Python版本

# 解法：
# 总的有效符号为 8 种 + - * ( ) 0 1 =
# 全排列枚举将有效符合去替换字母，通过 Python 的 eval 函数去判断等式左边 = 右边? (...我选择用 Python 过此题的唯一原因)
# 同时题面中还有部分限制规则需要判断。
# Python eval 计算二进制数需形如 eval('0b101+0b10') 。

lst = ['0', '1', '+', '-', '*', '(', ')', '=']
flg = [0 for i in range(8)]
ans = [0 for i in range(8)]
pos = [0 for i in range(256)]
chr = [0 for i in range(200)]
ok = list()
s = str()
cnt = 0
tot = [0]

def jug(t):
    for i in range(len(t)):
        if t[i] in '+*':
            if i == 0 or t[i-1] in '(+*-':
                return True

        elif t[i] == '0':
            if (i > 0 and t[i-1] not in '01') and (i+1 < len(t) and t[i+1] in '01'):
                return True
        elif t[i] == '(':
            if i+1 < len(t) and t[i+1] == ')':
                return True
    return False

def jugQ():
    tmp = s[:]
    t = s[:]
    for i in range(1, cnt+1):
        t = t.replace(chr[i], ans[i-1])

    while True:
        flag = False
        for i in range(1, len(t)):
            if (t[i-1] in '(+-*=') and (t[i] in '01'):
                t = t[:i] + 'b' + t[i:]
                flag = True
                break
        if flag == False:
            break
    if t[0] in '01':
        t = 'b' + t[:]

    t = t.replace('b', '0b')

    try:
        t1, t2 = t.split('=')
        if jug(t1) or jug(t2):
            return
        a1 = eval(t1)
        a2 = eval(t2)
        if a1 == a2 and t not in ok:
            ok.append(t)
            # print(tmp)
            # print(t)
            tot[0] += 1
    except:
        return

def dfs(idx):
    if idx == 8:
        # print(ans)
        jugQ()
    for i in range(8):
        if flg[i]:
            continue
        else:
            ans[idx] = lst[i]
            flg[i] = 1
            dfs(idx+1)
            flg[i] = 0


if __name__ == "__main__":

    s = input()
    for c in s:
        if (c >= 'a' and c <= 'z') or (c >= 'A' and c <= 'Z'):
            if pos[ord(c)] > 0:
                continue
            else:
                cnt+=1
                pos[ord(c)] = cnt
                chr[cnt] = c

    if cnt > 8:
        print(0)
    else:
        dfs(0)
        print(tot[0])