#include <iostream>
#include <stdio.h>
#include <iomanip>
#include <math.h>
using namespace std;
const int N = 1e6 + 10;
const int INF = 0x3f3f3f3f;
int main()
{
	int a;
	cin >> a;
	if( a % 3 == 0 && a % 5 == 0 )
	{
		cout << "YES";
	}
	else
	{
		cout << "NO";
	}
	return 0;
}

#include<iostream> using namespace std; int main() { int n; cin>>n; if(n%30&&n%50) { cout<<"YES"; } else{ cout<<"NO" ; } return 0; }

#include<iostream> using namespace std; int main(){ int n; cin>>n; if(n%30&&n%50) { cout<<"YES"; } else{ cout<<"NO" ; } return 0; }

#include <queue>
#include <math.h>
#include <stack>
#include <stdio.h>
#include <iostream>
#include <vector>
#include <iomanip>
#include <string.h>
#include <algorithm>
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
const int INF = 0x3f3f3f3f;
int main()
{
	int a;
	cin >> a;
	if( a % 3 == 0 && a % 5 == 0 )
	{
		cout << "YES";
	}
	else
	{
		cout << "NO";
	}
	return 0;
}

#include<iostream>

using namespace std;

int main()

{

int a;

cin>>a;

if(a%30&&a%50)

{

cout<<"YES";

}

else

{ cout<<"NO";

}

return 0;

}

可恶再度把添胜第3周没做完的愤怒发泄在题解上（？

最小公倍数指一个数能同时除以两个或以上的数，并且这个数是最小的，那这个数就是这些数 的最小公倍数。（不懂的话看五年级下册数学书去吧

这一题，能同时整除3和5的数也一定能整除3和5的最小公倍数15（15既能整除3又能整除5，而且15以内的正整数都不能同时整除3和5

上代码：

#include<bits/stdc++.h>
using namespace std;

int main(){
	int n;
	cin>>n;
	if(n%15==0)cout<<"YES";
	else cout<<"NO";
	return 0;
}

因为懒所以不加注释了，，

欢迎来luogu找我玩！

#include<iostream> using namespace std; int main(){ int a; cin>>a; if(a%30&&a%50){ cout<<"YES"; }else{ cout<<"NO"; } }

• #include <stdio.h>
  #include <iostream>
  #include <math.h>
  #include <iomanip>
  using namespace std;
  int main()
  {
  	int n;
  	cin>>n;
  	if(n % 3 == 0 && n % 5 == 0)
  	{
  		cout<<"YES";
  	}
  	else
  	{
  		cout<<"NO";
  	}
  }
  

  Copy *

#include <stdio.h>
#include <iostream>
#include <math.h>
#include <iomanip>
using namespace std;
int main()
{
	int n;
	cin>>n;
	if(n % 3 == 0 && n % 5 == 0)
	{
		cout<<"YES";
	}
	else
	{
		cout<<"NO";
	}
}

#include<bits/stdc++.h> using namespace std; int main() { int n; cin>>n; if(n%30&&n%50) cout<<"YES"; else cout<<"NO"; return 0; }

#include<iostream> #include<iomanip> using namespace std; int main(){ int a; cin>>a; if (a%30 and a%50){ cout<<"YES"; } else{ cout<<"NO"; } }

#include<iostream>

using namespace std;

int main(){

int a;

cin>>a;

if(a%15==0){

	cout<<"YES";

}else{

	cout<<"NO";

}

} //极简主义

#include<iostream> #include<stdio.h> using namespace std; int main() { long long a; cin>>a; if(a%30&&a%50) { cout<<"YES"<<endl; }else { cout<<"NO"<<endl; } }

#include <stdio.h>
#include <iostream>
#include <math.h>
#include <iomanip>
using namespace std;
int main()
{
	int n;
	cin>>n;
	if(n%3==0&&n%5==0)
	{
		cout<<"YES";
	}
	else
	{
		cout<<"NO";
	}
}
``

n = int(input()) if n % 3 == 0 and n % 5 == 0 and n != 0: print("YES") if n % 3 != 0 or n % 5 != 0 or n % 3 != 0 and n % 5 == 0 or n % 3 == 0 and n % 5 != 0 or n == 0: print("NO")