逆波兰计算器&&中缀表达式转换为后缀表达式

逆波兰计算器

#include
#define INF 0x3f3f3f3f3f3f3f3f
#define mod 1000000007
#define IOS ios::sync_with_stdio(false)
#define endl '\n'
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 10;
stack<double>stk;
int main() {
	IOS;
	double a, b;
	char c, str[10];
	int i = 0;
	scanf("%c", &c);
	while (c != '#') {
		while (isdigit(c) || c == '.') {
			str[i++] = c;
			str[i] = '\0';
			scanf("%c", &c);
			if (c == ' ') {
				a = atof(str);
				stk.push(a);
				i = 0;
				break;
			}
		}
		switch (c) {
		case '+': {
			a = stk.top();
			stk.pop();
			b = stk.top();
			stk.pop();
			stk.push(a + b);
			break;
		}
		case'-': {
			a = stk.top();
			stk.pop();
			b = stk.top();
			stk.pop();
			stk.push(b - a);
			break;
		}
		case'*': {
			a = stk.top();
			stk.pop();
			b = stk.top();
			stk.pop();
			stk.push(a * b);
			break;
		}
		case'/': {
			a = stk.top();
			stk.pop();
			b = stk.top();
			stk.pop();
			if (a == 0) {
				printf("Error!除数不能为0");
				return -1;
			}
			stk.push(b / a);
			break;
		}
		}
		scanf("%c", &c);
	}
	printf("%llf", stk.top());
	return 0;
}

输入

1 2 3 - 4 * + 10 5 / + # (以#为结束符)

输出

-1.000000

中缀表达式—>后缀表达式

#include
#define INF 0x3f3f3f3f3f3f3f3f
#define mod 1000000007
#define IOS ios::sync_with_stdio(false)
#define endl '\n'
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 10;
stack<char>s;
string ss;
int getnum(int& i,bool negtive) {
	int ans = 0;
	while (ss[i]>='0' && ss[i]<='9') {
		ans = ans * 10 + ss[i] - '0';
		++i;
	}
	if (negtive)ans = -ans;
	return ans;
}
int main() {
	char c;
	char temp;
	getline(cin, ss);
	int i;
	for (i = 0;i < ss.size();++i) {
		while (isdigit(ss[i])) {
			printf("%c", ss[i]);
			++i;
			if (ss[i]<'0' || ss[i]>'9') {
				printf(" ");
				break;
			}
		}
		if (ss[i] == ')') {
			do {
				temp = s.top();
				s.pop();
				if (temp != '(')printf("%c ", temp);
			} while (temp != '(');
		}
		else if (ss[i] == '+' || ss[i] == '-') {
			if (s.empty())s.push(ss[i]);
			else {
				do {
					temp = s.top();
					s.pop();
					if (temp == '(')s.push(temp);
					else {
						printf("%c ", temp);
					}
				} while (!s.empty() && temp != '(');
				s.push(ss[i]);
			}
		}
		else if (ss[i] == '*' || ss[i] == '/' || ss[i] == '(')s.push(ss[i]);
	}
	while (!s.empty()) {
		temp = s.top();
		s.pop();
		printf("%c ", temp);
	}
	return 0;
}

输入

1+(2-3)*4+10/5

输出

1 2 3 - 4 * + 10 5 / +