题解 | #牛牛的二三七整除#

知识点：

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#include <iostream>
using namespace std;

int main() {
    int num = 0;
    int count = 0;
    int arr[3] = {2, 3, 7};

    // scanf("%d", &num);
    cin >> num;

    for (int i = 0; i < 3; i++) {
        if (num % arr[i] == 0) {
            // printf("%d ", arr[i]);
            cout << arr[i] << " ";
            count++;
        }
    }

    if (count == 0) {
        // printf("n");
        cout << "n";
    }
    return 0;
}

// 原来解法
/*
int main() {
    int num = 0;
    int two = 2;
    int three = 3;
    int seven = 7;
    bool flag = true;

    cin >> num;

    if (num % two == 0) {
        cout << two << " ";
        flag = false;
    }

    if (num % three == 0) {
        cout << three << " ";
        flag = false;
    }

    if (num % seven == 0) {
        cout << seven;
        flag = false;
    }

    if (flag) {
        cout << "n";
    }

    return 0;
}
*/