Common divisor

#include <iostream>

using namespace std;

int main()

{

   int a,b,c[50],d[50],count=0;

   cin>>a>>b;

   for(int i=0;i<50;i++){

       if(a%(i+1)==0){

           c[i]=i+1;

       }

       

   }

    for(int i=0;i<50;i++){

       if(b%(i+1)==0){

           d[i]=i+1;

       }

       

   }

   for(int i=0;i<50;i++){

       if(c[i]==d[i]){

           cout<<c[i]<<endl;

       

   }}

  

}

Palindrome

#include<iostream>
#include<string>
using namespace std;
int main(){
string s;
cin>>s;
int b=0;
for(int i=0;i<s.size();i++){if(s[i]!=s[s.size()-i-1])

{b=1;
break;}
}
if(b==0){
    cout<<"yes";
}
else cout<<"N0";
    }
    

Graph Theory

Graph Theory

Adjacency matrix

An adjacency matrix is a VxV binary matrix A. Element $A_{i,j}$ is 1 if there is an edge from vertex i to vertex j else $A_{i,j}$ is 0.

Note: A binary matrix is a matrix in which the cells can have only one of two possible values - either a 0 or 1.

The adjacency matrix can also be modified for the weighted graph in which instead of storing 0 or 1 in $A_{i,j}$, the weight or cost of the edge will be stored.

In an undirected graph, if $A_{i,j}$ = 1, then $A_{j,i}$ = 1. In a directed graph, if $A_{i,j}$ = 1, then $A_{j,i}$ may or may not be 1.

Adjacency matrix provides constant time access (O(1) ) to determine if there is an edge between two nodes. Space complexity of the adjacency matrix is O($V^2$).

The adjacency matrix of the following graph is:

i/j: 1 2 3 4
1 : 0 1 0 0
2 : 0 0 0 1
3 : 1 0 0 1
4 : 0 1 0 0

code for adjacency matrix

#include<iostream>
using namespace std;
int main(){

    int x,y,node,edge;

    cin>>node>>edge;

    int a[10][10]={0};

    for(int i=0;i<edge;i++){

        cin>>x>>y;

        a[x][y]=1;
        
    }
  
}

TEST YOUR UNDERSTANDING

Edge Existence

You have been given an undirected graph consisting of $N$ nodes and $M$ edges. This graph can consist of self-loops as well as multiple edges. In addition , you have also been given $Q$ queries. For each query , you shall be given 2 integers $A$ and $B$. You just need to find if there exists an edge between node $A$ and node $B$. If yes, print "YES" (without quotes) else , print "NO"(without quotes).
Input Format:
The first line consist of 2 integers $N$ and $M$ denoting the number of nodes and edges respectively. Each of the next $M$ lines consist of 2 integers $A$ and $B$ denoting an undirected edge between node $A$ and $B$. The next line contains a single integer $Q$ denoting the number of queries. The next Line contains 2 integers $A$ and $B$ denoting the details of the query.
Output Format
Print Q lines, the answer to each query on a new line.
Constraints:
$1\le N\le {10}^3$
$1\le M\le {10}^3$
$1\le A,B\le N$
$1\le Q\le {10}^3$

#include<iostream>
using namespace std;
int main(){

    int x,y,node,edge;

    cin>>node>>edge;

    int a[10][10]={0};

    for(int i=0;i<edge;i++){

        cin>>x>>y;

        a[x][y]=1;
        
    }
  int r,p,q;
  cin>>r;
  for(int i=0;i<r;i++){
      cin>>p>>q;
      if(a[p][q]==1){
          cout<<"YES"<<endl;
          
      }
      else cout<<"NO"<<endl;
      
  }

}

Bubble Sorting

#include <iostream>

using namespace std;

void swap(int &a,int &b){

    int r;

    r=a;

    a=b;

    b=r;}

int main()

{

  int n;

  cin>>n;

  int a[n];

  for(int i=0;i<n;i++){

      cin>>a[i];

  }

  for(int i=0;i<n;i++){

      for(int j=0;j<n;j++){

          if(a[j]>a[j+1])

          swap(a[j],a[j+1]); }

  }

  for(int i=0;i<n;i++){

      cout<<a[i]<<endl;}

}

Lonely Integer

Consider an array of  integers, , where all but one of the integers occur in pairs. In other words, every element in  occurs exactly twice except for one unique element.

Given , find and print the unique element.

Input Format

The first line contains a single integer, , denoting the number of integers in the array. 
The second line contains  space-separated integers describing the respective values in .

Constraints

• It is guaranteed that  is an odd number.
• , where .

Output Format

Print the unique number that occurs only once in  on a new line.

Sample Input 0

1
1

Sample Output 0

1

Explanation 0 
The array only contains a single , so we print  as our answer.

Sample Input 1

3
1 1 2

Sample Output 1

2

Explanation 1 
We have two 's and one . We print , because that's the only unique element in the array.

Sample Input 2

5
0 0 1 2 1

Sample Output 2

2

Explanation 2 
We have two 's, two 's, and one . We print , because that's the only unique element in the array.

Solution:

#include<iostream>

using namespace std;

int main(){

int n;

cin>>n;

int a[1000]={0},b;

for(int i=0;i<n;i++){

    cin>>b;

    a[b]++;

}

for(int i=0;i<1000;i++){

if(a[i]==1)

    cout<<i;}

}