You just need to subtract the original massage from the key.

#include<iostream>
using namespace std;
int main(){
  int m,k;
  cin>>m>>k;
  cout<< m - k;
  return 0;
}

If x is greater than or equal b then it is clear that you need to add (x — b) boxes, If not then you needs to find how many full sets of boxes you can transform until you can't or you get zero boxes left, the best way to do this is to take the modulo.

#include<iostream>
using namespace std;
int main(){
  int a,b;
  cin>>a>>b;
  cout << (a - (b%a))%a <<endl;
  return 0;
}

Find the number of hours then the number of minutes then the number of seconds by using division and modulo.

#include<iostream>
using namespace std;
int main(){
  long long x;
  cin>>x;
  long long sec,min,hours;
  sec = x % 60;
  min = (x / 60) % 60;
  hours = x / (60 * 60);
  cout<<hours<<":"<<min<<":"<<sec;
  return 0;
}

Since each level consists of 2 normal-mobs waves and 1 boss fight, then the time needed to pass one level is (2 * m + b).

#include<iostream>
using namespace std;
int main(){
  int t,m,b;
  cin>>t>>m>>b;
  t*=60;
  int one_lvl_cost = 2*m + b;
  int ans = t / one_lvl_cost;
  cout<<ans;
  return 0;
}

Find the rate at which Ali and Yahya plant trees per 1 unit minute which (how many trees you plant divided by how much time is needed), then divide how many trees you want to plant by this rate.

#include<iostream>
#include<iomanip>
using namespace std;
int main(){
  int x,y,a,b;
  int l;
  cin>>x>>y>>a>>b>>l;
  float ali_rate = float(x) / a;
  float yahya_rate = float(y) / b;
  float ans = l / (yahya_rate + ali_rate);
  cout<<fixed<<setprecision(6)<<ans;
  return 0;
}

First let's find the link between whether a number is even or odd and the gcd of this number with another number, gcd (greatest common divisor) is just a divisor and we know that an odd number cant have an even divisor So we can't count any pair which have an odd number.

Now we are left with only the even numbers from 1 to N let's call it E and it's equal to (N/2) floored.

We have a last step which is to count how many distinct pairs I can make out of E numbers, for example, if I have four numbers (2, 4, 6, 8) I can make six distinct pairs which are: {(2,4) , (2,6) , (2,8) , (4,6) , (4,8) , (6,8)}, if you look closely we are summing up numbers from 1 to N-1 to find the number of pairs like this (3 + 2 + 1) and if we had any other number we would do the same thing.

So now we only need to find a way to sum up numbers from 1 to N-1, simply there is a formula to calculate it and we call it the summation formula which is: E * (E+1) / 2.

This is an explanation of the proof for this formula if you are interested in it : Gausse proof

#include<iostream>
#include<iomanip>
using namespace std;
int main(){
  int n; cin>>n;
  long long evens = n / 2; //integer division
  evens -= 1; // becuase we sum up numbers from 1 to evens-1
  long long ans = (evens) * (evens + 1) / 2;
  cout<< ans;
  return 0;
}