+-*/()整型運算式求值（不帶符號）

map<char, int> opt;double cal(double lv, double rv, char op) {switch (op) {case '+': return lv + rv;case '-': return lv - rv;case '*': return lv * rv;case '/': return lv / rv;}assert(0);}double exp_cal(const string &str) {int sz = str.size();vector<int> mp(sz, -1);stack<int> lp;for (int i = 0; i < sz; ++i) {if (str[i] == '(') {lp.push(i);} else if (str[i] == ')') {mp[i] = lp.top();mp[lp.top()] = i;lp.pop();}}stack<double> e, op;double v = 0;for (int i = 0; i <= sz; ++i) {if (isdigit(str[i])) {v *= 10;v += str[i] - '0';} else {if (str[i] == '(') {string tmp(str.begin() + i + 1, str.begin() + mp[i]);v = exp_cal(tmp);i = mp[i];} else {e.push(v);v = 0;while (!op.empty() && opt[str[i]] <= opt[op.top()]) {double second = e.top(); e.pop();double first = e.top(); e.pop();char operation = op.top(); op.pop();e.push(cal(first, second, operation));}op.push(str[i]);}}}assert(e.size() == 1 && op.size() == 1);return e.top();}int main(){opt[0x0] = 0;opt['+'] = 1;opt['-'] = 1;opt['*'] = 2;opt['/'] = 2;cout << exp_cal("3") << endl;cout << exp_cal("2+3") << endl;cout << exp_cal("2*3") << endl;cout << exp_cal("2+3*4") << endl;cout << exp_cal("2*3+4") << endl;cout << exp_cal("2+3+4") << endl;cout << exp_cal("2-3-4") << endl;cout << exp_cal("2+3*4-5") << endl;cout << exp_cal("2*3/4*5") << endl;cout << endl;cout << exp_cal("2*(3*(4+5))+2*(2+4)") << endl;cout << exp_cal("(10/(2+3)*(5-3))") << endl;system("pause");return 0;}