Codeforces Problem 864A Solution in (C++)

Welcome:

The question to the problem can be found here:

Explanation:
1. a[entered input] is the location of the input, say 10 it means the element is in the 10th position in the array.
2. Increase the value in that position by 1 when you get a different match of integers.
3. Store the first identical element in the 1st place of another array com[0] , second element to com[1]
4. Print YES if there are at exactly 2 different number in all cases.Then print those two numbers.
5. If not Print NO
6. Happy Coding!

Code:
#include<bits/stdc++.h>
using namespace std;
int a[100], com[2];
main(){
    int sto=0;
int n;
cin>>n;
int x;
for(int i=0;i<n;i++){
cin>>x;
cin.ignore();
if(++a[x] == 1){ //if more than 1 loop condition becomes false;
    com[sto++] = x;
  
}
}

if(sto == 2 && a[com[0]]== a[com[1]]){
    cout<<"YES"<<endl;
    cout<<com[0]<<" "<<com[1];
}
else{
    cout<<"NO"<<endl;
}

}

CodeForces Problem 862A - Mahmoud and Ehab and the MEX Solution (C++)

Problem:

Please find the problem here.

Solution:

This is the simplest problem I ever had, just implement the given formulas! However, it does take me some time to write.

Explanation:

1.MEX has two condition

2.MEX equal to entered input ... MEX will increase by 1 as element is erased.

3.MEX less than input...MEX will decline as element is inserted before MEX

4.Goodluck!

Code:

#include<bits/stdc++.h>

#define FOR(i,n) for(int i=0;i<n;i++)

using namespace std;

main(){

int n;

int k;

int x;

cin>>n>>k;

int ans=k;

FOR(i,n){

cin>>x;

if(x<k){

ans--;

}

else if(x==k){

ans++;

}

}

cout<<ans;

}

Uva Problem 11936 - The Lazy Lumberjacks Solution (C++)

Problem:

11936 - The Lazy Lumberjacks

Please find the problem here.

Solution:

This is the simplest problem I ever had, just implement the given formulas! However, it does take me some time to write.

Code:

#include <bits/stdc++.h>

using namespace std;

int main(){

int t;

int x,y,z;

cin>>t;

while(t--){

cin>>x>>y>>z;

if(x+y>z && y+z>x && z+x>y){

cout<<"OK"<<endl;

}

else if(x==y && y ==z && z==x)

    cout<<"OK"<<endl;

else cout<<"Wrong!!"<<endl;

}

return 0;}