思路

边取边模。

时间复杂度:

$O(n)$

空间复杂度:

$O(1)$

Code

#include<bits/stdc++.h>
using namespace std;
int n,m=1;
int main(){
	cin>>n;
	if(n==0){cout<<"YES";return 0;}
	for(int i=1;i<=n;i++){
		m*=i;
		m=m%(n+1);
	}
	if(m==0) cout<<"YES";
	else cout<<"NO";
	return 0;
}

思路

暴力出奇迹，打表出省一

解题方法

一边判断一边走

复杂度

全是判断包是O(1)

时间复杂度:$O(1)$

空间复杂度: $O(1)$

Code

#include<bits/stdc++.h>
using namespace std;
int main(){
    int a;
    cin>>a;
    if(a==1)cout<<"NO";
    else if(a==2)cout<<"NO";
    else if(a==3)cout<<"NO";
    else if(a==4)cout<<"NO";
    else if(a==6)cout<<"NO";
    else if(a==10)cout<<"NO";
    else if(a==12)cout<<"NO";
    else if(a==16)cout<<"NO";
    else if(a==18)cout<<"NO";
    else cout<<"YES";
    return 0;
}

思路

打表。

提前算好。

时间复杂度:

$O(1)$

空间复杂度:

$O(1)$

Code

#include<bits/stdc++.h>
using namespace std;
string s[20]={"YES","NO","NO","NO","NO",
"YES","NO","YES","YES","YES",
"NO","YES","NO","YES","YES",
"YES","NO","YES","NO","YES"};
int main()
{
	int n;
	cin>>n;
	cout<<s[n]<<endl;
	return 0;
}

O(1)水过

思路

暴力出奇迹，打表出省一

解题方法

将0！~20！依次除1~21

复杂度

时间复杂度:

$O(1)$

空间复杂度:

$O(1)$

Code

#include<bits/stdc++.h>
using namespace std;
string s[20]={"YES","NO","NO","NO","NO",
"YES","NO","YES","YES","YES",
"NO","YES","NO","YES","YES",
"YES","NO","YES","NO","YES"};
int main()
{
	int n;
	cin>>n;
	cout<<s[n]<<endl;
	return 0;
}