Skip to main content

URI ONLINE JUDGE -1044

Solution of 1044 number !

please do not copy it rather watch it for learning purpose :)




#include <iostream>
#include <bits/stdc++.h>
using namespace std;
int main ()
{
    int a,b;
    cin >> a >> b;
    if(a<b)
    {
    if(b%a==0 ) cout << "Sao Multiplos" << endl;
    else     cout << "Nao sao Multiplos" << endl;
    }
    else
    {
        if(a%b==0 ) cout << "Sao Multiplos" << endl;
    else     cout << "Nao sao Multiplos" << endl;
    }
    return 0;
}






If you do not understand please leave a message here.

or mail us mechaboo786@gmail.com

or you can leave a message at fb . Link :- https://www.facebook.com/mechaboo/

Comments

Popular posts from this blog

Skeleton Of C++

Today we will see the body structure of c++ below is given the structure:- #include <iostream> #include <bits/stdc++.h> using namespace std; int main () {     //your codes goes here     return 0; }   You will start writing your code from main function.

UVA-11764 Solution

Solution of 11764 UVA number ! please do not copy it rather watch it for learning purpose :) Here is the code :- #include <iostream> #include <bits/stdc++.h> using namespace std;    int main(){        //mechaboo     int testCases;     cin >> testCases;     for (int i=0;i<testCases;i++)     {         int len,maxx=0,minn=0;         cin >> len;         int  arrJumps[len] ;         for(int j=0;j<len;j++)         {             cin >> arrJumps[j];             if(j!=0)             {       ...

10424 - Love Calculator UVA Solution

10424 - Love Calculator UVA Solution please do not copy it rather watch it for learning purpose :) Here is the code :-  #include <iostream> #include <bits/stdc++.h> #include <string> using namespace std; int getval(string nam); int reducingVal(int val);    int main(){        //MechaBoo     string name1,name2;     while(getline(cin,name1))     {         getline(cin,name2);         int par1= getval(name1);         int par2=getval(name2);         par1=reducingVal(par1);         par2=reducingVal(par2);               double x= par1*1.00;         double y=par2*1.00;           ...