ZeroJudge K678: Adding Numbers

Feiyu successfully found his way out of the maze, but! Feiyu has no combat abilities, he can only go to the shop in Math City to purchase a wooden sword.

But wait! He's not in Math City anymore; he's in Anti-Math City.

Feiyu can only defeat Anti-Math Monsters to escape Anti-Math City!!!

肥余遇到在反數學城的第一隻怪獸是 —– 反加法獸!!!

Sample Inputs/Outputs

Sample Input(s)Sample Output(s)
The first line indicates the number of monsters (Anti-Addition Beasts).
Then repeat the following procedure the number of times indicated in the first line.
Input an addition equation.
Also, input the answer.		If the answer is correct, output "yes". If the answer is wrong, output "no".
3
1+1
2
2+4
5
5+8
4		yes
no
no

Thought Process

Use a string to store the addition expression. Run a for loop over the string. Declare an empty string, and concatenate each character to it in each iteration. If the current character is "+", convert the accumulated string into an integer and add it to the integer obtained from string.substr(current position + 1, string.length()).

If the calculated answer matches the input, output "yes"; otherwise, output "no".

Sample Code－ZeroJudge K678: 肥余歷險記—(反數學城1)

#include <iostream>
using namespace std;

int toint(string str)
{
    if (str.length() == 1) return int(str[0] - '0');
    return stoi(str);
}

int main() {
    cin.sync_with_stdio(0);
    cin.tie(0);
    int N;
    cin >> N;
    for (int i = 0; i<N; i++)
    {
        string str, tmp = "";
        cin >> str;
        int sum = 0;
        for (int j = 0; j<str.length(); j++)
        {
            if (str[j] == '+')
            {
                sum += toint(tmp);
                sum += toint(str.substr(j+1, str.length()));
                break;
            }
            tmp += str[j];
        }
        int ans;
        cin >> ans;
        if (sum == ans) cout << "yes\n";
        else cout << "no\n";
    }
}

//ZeroJudge K678
//Dr. SeanXD
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