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<bits/stdc++.h>
using namespace std;
#define Result pair<char*,int>
#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])