表达式求值

#include <bits/stdc++.h>

using namespace std;

char OprateRelation[7][7] =// +    -    *    /    (    )   =
                            {{‘>‘,‘>‘, ‘<‘, ‘<‘, ‘<‘, ‘>‘, ‘>‘}, //‘+‘
                            {‘>‘, ‘>‘, ‘<‘, ‘<‘, ‘<‘, ‘>‘, ‘>‘}, //‘-‘
                            {‘>‘, ‘>‘, ‘>‘, ‘>‘, ‘<‘, ‘>‘, ‘>‘}, //‘*‘
                            {‘>‘, ‘>‘, ‘>‘, ‘>‘, ‘<‘, ‘>‘, ‘>‘}, //‘/‘
                            {‘<‘, ‘<‘, ‘<‘, ‘<‘, ‘<‘, ‘=‘, ‘ ‘}, //‘(‘
                            {‘>‘, ‘>‘, ‘>‘, ‘>‘, ‘ ‘, ‘>‘, ‘>‘}, //‘)‘
                            {‘<‘, ‘<‘, ‘<‘, ‘<‘, ‘<‘, ‘ ‘, ‘=‘}};//‘=‘

string Operteraiton = "+-*/()=";

int Getoperteration(char x) {
    for(int i = 0; i < 7; i++)
        if(Operteraiton[i] == x) return i;
    return -1;
}
char CompareOperation(char x, char y) {
    return OprateRelation[Getoperteration(x)][Getoperteration(y)];
}
double Get_ans(double x, double y, char opt) {
    if(opt == ‘-‘) return x - y;
    else if(opt == ‘+‘) return x + y;
    else if(opt == ‘*‘) return x * y;
    else return x/y;
}
double Slove(string Pattern) {
    stack <double> Num;
    stack <char> Opertion;
    Opertion.push(‘=‘);
    int i = 0, flag = 1, gg = 1;
    char Next_char;
    if(Pattern[0] == ‘+‘ || Pattern[0] == ‘-‘) Num.push(0);
    if(Pattern[Pattern.size() - 1] != ‘=‘) Pattern += ‘=‘;
    while(flag && i < Pattern.size()) {
        Next_char = Pattern[i];
        if(Getoperteration(Next_char) < 0) {
            string Number = "";
            Number += Next_char;
            while(Getoperteration(Pattern[++i]) < 0) {
                Number += Pattern[i];
            }
            Num.push(atof(Number.c_str()));
        }else {
            char opt = CompareOperation(Opertion.top(), Next_char);
            if(opt == ‘<‘) {
                Opertion.push(Next_char);
                i++;
            }else if(opt == ‘>‘) {
                char Next_option = Opertion.top();
                Opertion.pop();
                double y = Num.top();
                Num.pop();
                double x = Num.top();
                Num.pop();
                double ans = Get_ans(x, y, Next_option);
                Num.push(ans);
            }else {
                Opertion.pop();
                if(i == (int)Pattern.length()) flag = 0;
                else i++;
            }
        }
    }
    return Num.top();
}
int main()
{
    int T;
    cin >> T;//测试用例的个数
    string s;
    while(T--) {
        cin >> s;
        printf("%.2f\n", Slove(s));
    }
    return 0;
}