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 Trueelif 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 Trueelif t[i] == '(':if i+1 < len(t) and t[i+1] == ')':return Truereturn Falsedef jugQ():tmp = s[:]t = s[:]for i in range(1, cnt+1):t = t.replace(chr[i], ans[i-1])while True:flag = Falsefor i in range(1, len(t)):if (t[i-1] in '(+-*=') and (t[i] in '01'):t = t[:i] + 'b' + t[i:]flag = Truebreakif flag == False:breakif t[0] in '01':t = 'b' + t[:]t = t.replace('b', '0b')try:t1, t2 = t.split('=')if jug(t1) or jug(t2):returna1 = eval(t1)a2 = eval(t2)if a1 == a2 and t not in ok:ok.append(t)# print(tmp)# print(t)tot[0] += 1except:returndef dfs(idx):if idx == 8:# print(ans)jugQ()for i in range(8):if flg[i]:continueelse:ans[idx] = lst[i]flg[i] = 1dfs(idx+1)flg[i] = 0if __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:continueelse:cnt+=1pos[ord(c)] = cntchr[cnt] = cif cnt > 8:print(0)else:dfs(0)print(tot[0])

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

AC代码

C++版本

Python版本

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