【C++ | PTA】Easy chemistry

题目要求

In this question, you need to write a simple program to determine if the given chemical equation is balanced. Balanced means that the amount of elements on both sides of the “=” sign is the same.

$$H_{2}O+C{O}_2=H_{2}C{O}_3$$
We guarantees that each chemical equation satisfies the basic rules.

If you don’t know the basic rules of chemical equation writing, you can refer to the following conditions:

• The chemical equation contains exactly one “=” sign.
• For every chemical element, the first letter of each element is uppercase, and the rest of the letters are lowercase.
• The number at the end of the element represents the amount of this element needed, If there is no number, it means only one element is needed.
• Every chemical object may contain several chemical elements like NaCl.
• The number at the beginning of every chemical object represents the number of this object, If there is no number, it means only one object is needed.
• Every chemical object will be connected by “+” sign.
• Chemical reaction is considered to rearrange the atoms after they are broken up

Input
The first line contains a single integer T(1≤T≤100), indicating the number of test cases.
In the next T lines, each line contains a string S(1≤∣S∣≤100), representing a Chemical equation.
It is guaranteed that S only contains uppercase letters, lowercase letters, “=” sign and “+” sign, the sum of the amount of all elements will not exceed 2147483647.

Output
If the given chemical equation is balanced, please output “Easy!”, otherwise output “Hard!” (Without quotes).

Sample Input
3
H2O+CO2=H2CO3
2NaHCO3=Na2CO3+H2O+CO2
3Cu+8HNO3=3CuNO3+2NO+4H2O

Sample Output
Easy!
Easy!
Hard!

题目翻译

在这道题目中，你需要编写一个简单的程序来判断给定的化学方程式是否平衡。平衡指的是等号两侧所含各元素的数量相等。

例如，对于方程式 $H_{2}O+C{O}_2=H_{2}C{O}_3$，它是平衡的。

我们保证每个化学方程式都满足以下基本规则：

• 化学方程式中恰好有一个等号。
• 每个元素的第一个字母大写，其余字母小写。
• 元素后面的数字表示该元素需要的数量，如果没有数字，则表示只需要一个该元素。
• 每个化学物质可能包含多个元素，如 NaCl。
• 每个化学物质前面的数字表示该物质的个数，如果没有数字，则表示只需要一个该物质。
• 化学反应被认为是将原子打散重新组合。

输入格式：

第一行包含一个整数 T $(1\le T\le 100)$，表示测试用例的数量。
接下来有 T 行，每行包含一个字符串 S $(1\le |S|\le 100)$，表示一个化学方程式。
保证 S 只包含大写字母、小写字母、等号和加号，所有元素数量之和不会超过 $2^{31}-1$。

输出格式：

对于每个测试用例，如果其对应的化学方程式平衡，则输出 “Easy!”，否则输出 “Hard!”。

输入样例：
3
H2O+CO2=H2CO3
2NaHCO3=Na2CO3+H2O+CO2
3Cu+8HNO3=3CuNO3+2NO+4H2O

输出样例：
Easy!
Easy!
Hard!

代码

#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <map>
using namespace std;

map<string, int> m;

void GetElement(string str, int div) {
    string s0, s1;
    int coe = 1;
    int back = 1;

    int i = 0;
    while (i < str.size() && str[i] >= '0' && str[i] <= '9')
        s0 += str[i++];
    if (!s0.empty())
        coe = stoi(s0);
    s0.clear();
    for (; i < str.size(); i++) {
        if (str[i] >= 'A' && str[i] <= 'Z') {
            s0 += str[i];
            while (i + 1 < str.size() && str[i + 1] >= 'a' && str[i + 1] <= 'z')
                s0 += str[++i];

            int tag = 0;
            while (i + 1 < str.size() && str[i + 1] >= '0' && str[i + 1] <= '9')
                tag = 1, s1 += str[i + 1], i++;
            if (tag) back = stoi(s1), i--;

            if (div)
                m[s0] -= coe * back;
            else
                m[s0] += coe * back;
            s0.clear();
            s1.clear();
            back = 1;
        }
    }
}

bool isEqual(string str) {
    vector<string> v;
    string s0;
    for (int i = 0; i < str.size(); i++) {
        if (str[i] == '=') v.push_back("-1");
        int tag = 0;
        while (i < str.size() && str[i] != '+' && str[i] != '=') tag = 1, s0 += str[i], i++;
        if (tag) v.push_back(s0), i--;
        s0.clear();
    }

    int div = 0;
    for (int i = 0; i < v.size(); i++) {
        if (v[i] == "-1") {
            div = 1;
            continue;
        }
        GetElement(v[i], div);
    }

    for (auto it = m.begin(); it != m.end(); it++)
        if (it->second) return false;
    return true;
}

int main() {
    int T;
    string str;
    cin >> T;
    while (T--) {
        cin >> str;
        if (isEqual(str))
            cout << "Easy!" << endl;
        else
            cout << "Hard!" << endl;
        m.clear();
    }

    return 0;
}