Any weight that is even and greater than $$$2$$$ can be divided into two even numbers, say $$$2$$$ and $$$(w-2)$$$, And in other cases, splitting into two even weights is impossible.

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

int  main()
{
   int w;
   cin>>w;
   if(w%2==0 && w!=2)
   {
      cout<<"YES"<<endl;
   }
   else
   {
      cout<<"NO"<<endl;
   }
   return 0;
}

For this problem you just need to implement what it asks you. To be able to implement it you need to know about the "if" statement.

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

int main() {
    int t;
    cin >> t;
    for(int i=1; i<=t; ++i)
    {
        int x; cin >> x;
        if(x < 1400) cout << "Division 4\n";
        else if(x < 1600) cout << "Division 3\n";
        else if(x < 1900) cout << "Division 2\n";
        else cout << "Division 1\n";
    }
}

Consider the two edge cases: all the $$$z$$$ persons upvote or all the $$$z$$$ persons downvote. If the results are the same in these two cases, it is the answer. Otherwise, the result is uncertain.

#include <bits/stdc++.h>
using namespace std;
int  main()
{
   int x, y, z;
   cin>>x>>y>>z;
   if(x>(y+z))
   {
      cout<<"+"<<endl;
   } 
   else if(y>(x+z))
   {
      cout<<"-"<<endl;
   }
   else if(x==y && z==0)
   {
      cout<<"0"<<endl;
   }
   else
   {
      cout<<"?"<<endl;
   }
}

In this problem just multiply $$$a$$$ by $$$3$$$ and $$$b$$$ by $$$2$$$ until $$$a$$$ > $$$b$$$. Output the number of operations.

#include<bits/stdc++.h>
using namespace std;
 
 
int main()
{
    int a, b, i;
    cin>>a>>b;
    for(i=0; a<=b ;++i)
    {
        a=3*a;
        b=2*b;
    }
    cout<<i;
    return 0;
}

We need to make sequence of moves like: 1, 2, 1, 2, ...

So the answer is $$$2 * n / 3$$$. After that we have either 0, 1 or 2 stones left. If we have 0, we are done, otherwise we have 1 or 2 left, so we only can give 1 more stone.

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

int  main()
{
   int n;
   cin>>n;
   if(n % 3 == 0)
   {
      cout<<2 * (n / 3);
   }
   else
   {
      cout<<2 * (n / 3) + 1;
   }
}

We can iterate through the values of $$$a$$$ from $$$1$$$ to n. For each $$$i$$$ if $$$n$$$ is divisible by $$$i$$$ and if difference $$$(n/i$$$ — $$$i)$$$ is less than already found difference we need to update answer with values $$$i$$$ and $$$n/i$$$ (because $$$a$$$ must be less than $$$b$$$).

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

int  main()
{
   int n;
   cin>>n;
   int a = 1;
   int b = n;
   int difference = (n-1);
   for(int i=1; i<=n/2; ++i)
   {
      if(n%i==0)
      {
         if(i > n/i)
         { 
            continue;
         }
         if(n/i - i < difference)
         {
            a = i;
            b = n/i;
            difference = n/i - i;
         }
      }
   }
   cout<<a<<" "<<b<<endl;
}