不正经题解 | 判断整除

温馨提示
该贴为双语言贴，下方有英文版。
Warm reminder
This post is bilingual, with an English version below.

思路分析
根据题意，我们可以知道其实这一道题实际上就是让我们求出这个数是不是3或5的倍数。我们可以直接通过%来求出是否是等于0，也可以通过创建一个dable类型的变量，算出n分别取除以3和5的数值，在通过%1，和dable变量的储存来判断他是否是3和5的整数。
答案

#include <iostream>
using namespace std;
bool daan(int n){
    if(n % 3 == 0 && n % 5 == 0)
        return true;
    else
        return false;
}
int main(){
    int n;
    cin >> n;
    if(daan(n))
        cout << "YES";
    else 
        cout << "NO";
    return 0;
}

答案讲解
在这里呢，我运用到了函数，通过函数来判断是否是3或5的倍数，如果满足要求则返回true，不满足返回false，在通过if的判断来输出
如果是true，则输出“YES”
否则输出“NO”
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我有一个问题：
名字：
问题：

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Train of thought analysis
According to the meaning of the question, we can know that this question is actually asking us to find out if this number is * multiple of 3 or 5. We can directly use% to determine whether it is equal to 0, or we can create * variable of type dab and calculate n by dividing it by 3 and 5, and then use the storage of% 1 and dab variables to determine whether it is an integer of 3 and 5.
Answer:

#include <iostream>
using namespace std;
bool daan(int n){
if(n % 3 == 0 && n % 5 == 0)
return true;
else
return false;
}
int main(){
int n;
cin >> n;
if(daan(n))
cout << "YES";
else 
cout << "NO";
return 0;
}

Answer explanation
Here, I applied * function to determine whether it is * multiple of 3 or 5. If it meets the requirements, it returns true; if it does not, it returns false. The output is determined by the if function
If true, output 'YES'
Otherwise, output 'NO'
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name:
Question:

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