题解：P14305 【MX-J27-T2】转换

#include <bits/stdc++.h>
using namespace std;
string add_mul(string a, string b) { //计算
	if (a == "double" || b == "double") {
		return "double";
	}
	if (a == "float" || b == "float") {
		return "float";
	}
	if (a == "longlong" || b == "longlong") {
		return "longlong";
	}
	return "int";
}
int main() {
	int c, t;
	cin >> c >> t;
	while (t--) {
		string s;
		cin >> s;
		string k = "";
		for (int i = 0; i < s.size(); i++) { //去除 ,
			if (s[i] == ',') {
				k = "";
				continue;
			}
			k += s[i];
		}
		vector <int> mul;
		vector <string> a;
		string tmp = "";
		int cnt = 0;
		for (int i = 0; i < k.size(); i++) { //分离符号和类型
			if (k[i] == '+' || k[i] == '*') {
				a.push_back(tmp);
				tmp = "";
				if (k[i] == '*') {
					mul.push_back(cnt); //只用保留 * 的位置，其余的默认为 +
				}
				cnt++;
			} else {
				tmp += k[i];
			}
		}
		a.push_back(tmp);
		vector <string> b;
		int start = 0;
		for (int i = 0; i < mul.size(); i++) { //去除 *
			for (int j = start; j < mul[i]; j++) { //把不用计算的存起来
				b.push_back(a[j]);
			}
			start = mul[i] + 2; //记得下标
			b.push_back(add_mul(a[mul[i]], a[mul[i] + 1]));
		}
		for (int i = start; i < a.size(); i++) {
			b.push_back(a[i]);
		}
		string ans = b[0];
		for (int i = 1; i < b.size(); i++) {
			ans = add_mul(ans, b[i]); //计算最终答案
		}
		cout << ans << endl;
	}
	return 0;
}