思路

用 a&1 可以保留 $a$ 二进制下的最后一位，从而判断奇偶性。

code

#include<bits/stdc++.h>
using namespace std;
int main(){
   int a;
   cin>>a;
   if(a&1)  cout<<"NO";
   else  cout<<"YES";
   return 0;
}

#include<bits/stdc++.h> using namespace std; int main() { int a; cin>>a; if(a%2==0) { cout<<"YES"; } else { cout<<"NO"; } return 0; }

#include<bits/stdc++.h> using namespace std; int main() { int a; cin>>a; if(a%2==0) { cout<<"YES"<<endl; } else { cout<<"NO"<<endl; } return 0; }

#include<iostream> using namespace std; int main() { int a,b; cin>>a; b=a%2; if(b==0) { cout<<"YES"<<endl; } else { cout<<"NO"<<endl; } return 0; }

#include<iostream> using namespace std; int main() { int a; cin>>a; if(a%2==0) { cout<<"YES"<<endl; } else { cout<<"NO"<<endl; } return 0; }

#include<bits/stdc++.h> using namespace std; int main() { int a; cin>>a; if(a%2==0) { cout<<"YES"; } else { cout<<"NO"; } return 0 ; }

==表判断左右是否相等，=表赋值

#include<bits/stdc++.h> using namespace std; int main() { int a; cin>>a; if(a % 2==0) { cout<<"YES"; } else { cout<<"NO"; } return 0; }

#include<bits/stdc++.h> using namespace std; int main() { int a; cin>>a; if(a%2==0) {cout<<"YES";} else{cout<<"NO";} return 0; }

YYDS

#include<bits/stdc++.h> using namespace std; int main() { int n; cin>>n; if(n%2==0) { cout<<"Yes"; } else { cout<<"No"; } return 0; }