C++实现带括号正整数加减乘除的计算器（代码中提供了测试用例）

100*(2+12)-(20/3)*2 100*(2-12)+(20/3)*(1-2) 10*(3+5*(5-5+1)*100+(5-9*(20-2)/2)*2*20/10)+1000 2/3-10 2+(2*4) 50/(50-20*2+1-9)*(10+12*12-12*10) 50*(50-20*2+1-9)*(10+12*12-12*10)/(1*2*3*4*5*6) (1)+(1+2)+2*(1+3) 100*(2+12)-20/3*2 7*(8+2-(8/2+5*4)) 3+((12+7*9)-15) ((5*(8+7)-5)+30*(7-6/3))*7 8/4*2 8/2*4 #include <iostream> #include <cstring> #include <cstdio> #include <stack> using namespace std; void step(stack<int>& num, stack<char>& op) { char top = op.top(); op.pop(); int a, b; b = num.top(); num.pop(); a = num.top(); num.pop(); //cout << a << top << b << "=";// if (top == '+') a += b; else if (top == '-') a -= b; else if (top == '*') a *= b; else a /= b; //cout << a << endl;// num.push(a); } bool cmp(char top, char c) { if (top == '(') return false; if (top == '+' || top == '-') { if (c == '+' || c == '-') return true; return false; } return true; } void pushOp(stack<int>& num, stack<char>& op, char c) { if (c == ')') { for (; op.top() != '(';) step(num, op); op.pop(); return; } if (op.empty() || op.top() == '(') { //cout << "New: " << c << endl;// op.push(c); return; } //cout << "Top: " << op.top() << " >> " << "New: " << c << endl;// for (; cmp(op.top(), c);) { //For the order of calculating step(num, op); if (op.empty()) break; } op.push(c); } int getNum(char* line, int& i, int len) { int a; sscanf(line + i, "%d", &a); //cout << a << endl;// for (; i < len && line[i] >= '0' && line[i] <= '9'; i++); return a; } int main() { stack<int> num; stack<char> op; char line[101]; int i, len; for (; scanf("%s", line) != EOF;) { i = 0; len = strlen(line); for (; line[i] == '('; i++) //I's like to get many '(' continously op.push('('); num.push(getNum(line, i, len)); for (; i < len;) { for (; line[i] == ')'; i++) //It's the same with '(' pushOp(num, op, line[i]); if (i >= len) break; pushOp(num, op, line[i++]); for (; line[i] == '('; i++) //... op.push('('); num.push(getNum(line, i, len)); } for (; !op.empty();) step(num, op); i = num.top(); num.pop(); //cout << "Empty: " << num.size() << endl; printf("%d\n", i); } return 0; }