tyvj/joyoi 1043 表达式计算4

这题怎么这么毒瘤...

一开始我想转后缀表达式来计算，后来发现有负数...弃疗。

递归求解又发现会有多余括号，我觉得不行...

实在是毒瘤啊！

 #include <cstdio>
 #include <iostream>
 #include <string>
 typedef long long LL;
 std::string s;

 bool isop(char c) {
     if(c == '+') return ;
     if(c == '-') return ;
     if(c == '*') return ;
     if(c == '/') return ;
     if(c == '^') return ;
     return ;
 }

 LL pow(LL a, LL b) {
     LL ans = ;
     for(int i = ; i <= b; i++) {
         ans *= a;
     }
     return ans;
 }

 LL solve(int l, int r) {
     int f = ;
     int pos1 = -, pos2 = -, pos3 = -;
     for(int i = l; i <= r; i++) {
         if(isop(s[i]) && (!f)) {
             if(s[i] == '+' || s[i] == '-') {
                 pos1 = i;
             }
             else if(s[i] == '*' || s[i] == '/') {
                 pos2 = i;
             }
             else {
                 pos3 = i;
             }
         }
         else if(s[i] == '(') f++;
         else if(s[i] == ')') f--;
     }

     if(f > ) {
         LL temp = solve(l + , r);
         return temp;
     }

     if(f < ) {
         LL temp = solve(l, r - );
         return temp;
     }

     if(pos1 > l) {
         if(s[pos1] == '+') {
             return solve(l, pos1 - ) + solve(pos1 + , r);
         }
         else {
             return solve(l, pos1 - ) - solve(pos1 + , r);
         }
     }
     else if(pos2 != -) {
         if(s[pos2] == '*') {
             return solve(l, pos2 - ) * solve(pos2 + , r);
         }
         else {
             return solve(l, pos2 - ) / solve(pos2 + , r);
         }
     }
     else if(pos3 != -) {
         return pow(solve(l, pos3 - ), solve(pos3 + , r));
     }
     else if(s[l] == '(' && s[r] == ')'){
         return solve(l + , r - );
     }
     else {
         LL ans = ;
         int f = ;
         while(s[l] == '(') l++;
         if(s[l] == '-') {
             f = -;
             l++;
         }
         for(int i = l; i <= r; i++) {
             if(s[i] == ')') break;
             ans = ans *  + s[i] - '';
         }
         return ans * f;
     }
 }

 int main() {
     std::cin >> s;
     printf("%lld", solve(, s.size() - ));
     return ;
 }

AC代码