decimel to ternary

DECIMEL TO TERNARY

#include<bits/stdc++.h>

using namespace std;

void bse(int n)

{

    if(n==0)return ;

    bse(n/3);

    printf("%d",n%3);

}

int main()

{

    int num;

    while(scanf("%d",&num)==1 && num>=0)

    {

        if(num==0)

        {

            printf("0\n");

            continue;

        }

        bse(num);

        printf("\n");

    }

}

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UVA 382

UVA 382

problem link: LINK

#include<bits/stdc++.h>
using namespace std;
void fnc(int num)
{
    int i,sum;
    sum=0;
    for(i=1;i<num;i++)
    {
        ///sum=0;
        if(num%i==0)
        {
            sum=sum+i;
        }
    }
    if(sum==num)
    {
        printf("%5d  PERFECT\n",num);
    }
    else if(sum>num)
    {
        printf("%5d  ABUNDANT\n",num);
    }
    else
    {
        printf("%5d  DEFICIENT\n",num);
    }

}
int main()
{
   int  ara[101],k,i;
   cout<<"PERFECTION OUTPUT"<<endl;
   for(i=0;i<=100;i++)
   {
       cin>>ara[i];
       k=i;
       if(ara[i]==0)
        break;

   }

   for(i=0;i<k;i++)
   {
       fnc(ara[i]);
   }
   cout<<"END OF OUTPUT"<<endl;




}

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The 3n+1 uva 100

The 3n+1 problem solution :

uva 100

problem link: Link

#include<bits/stdc++.h>
using namespace std;
int main()
{
    long long a,m,b,i,s,k,j,x;
    while(scanf("%lld%lld",&a,&b)==2){
   //cin>>a>>b;
    //cout<<a<<" "<<b;

       cout<<a<<" "<<b;
       if(a>b)
   {
       x=a;
       a=b;
       b=x;
   }



    //s=b;
    j=1;
    k=0;
    i=a;
    while(i<=b)

    {
        m=i;
        //cout<<i;
        while(m!=1)
        {
            if(m%2==0)
              {


                m=m/2;
                j=j+1;}
            else
                {m=3*m+1;
                j=j+1;}
        }



        //i=i+1;

        if(k<j)
        {
            k=j;

        }
        j=1;
        i=i+1;
    }

    cout<<" "<<k<<endl;}

    return 0;
}

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UVA 10696

 UVA 10696 solving in C++

 problem link:  link
#include<bits/stdc++.h>
using namespace std;
int main()
{
    unsigned long long gcd,lcm,y,x,g,m,n,l,t,i;
    while(scanf("%lld",&n)==1)
    {
        if(n==0)
            break;
        else if(n<=100)
        {
            printf("f91(%lld) = 91\n",n);
        }
        else
        {
            printf("f91(%lld) = %lld\n",n,n-10);
        }
    }

}

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uva 11388

 UVA 11388 solution in C++

problem link: Link

#include<bits/stdc++.h>
using namespace std;
int main()
{
    unsigned long long gcd,lcm,y,x,g,m,n,l,t,i;
    cin>>t;
    for(i=0;i<t;i++)
    {
        cin>>g>>l;
        y=max(g,l);
        x=(g+l)-y;
        ///cout<<y<<endl<<x;
        while(y%x!=0)
        {
            m=y;
            n=x;
            x=(m%n);
            y=n;


        }
        gcd=x;
        lcm=(g*l)/gcd;
        if((gcd!=g)&&(lcm!=l))
        {
            cout<<"-1"<<endl;
        }
        else
        {
            cout<<g<<" "<<l<<endl;
        }
    }
return 0;

}

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First post of my new blog

1. Problem: Watermelon

2. 
   
   

http://codeforces.com/problemset/problem/4/A

codeforces 4A(watermelon) solving in C++

Using modulus operator: 

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int a;
    cin>>a;
    if(a==2)
    {
        cout<<"NO";

   
    else if(a%2==0)
    {
        cout<<"YES";
    }
    else
    {
        cout<<"NO";
    } return 0; } 

 Using bitwise operator:

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int a=1,b,c;
    cin>>b;
    if(b==2)
    {
        cout<<"NO"<<endl;
    }
    else
    {
        c=a&b;
        if(c==0)
        {
            cout<<"YES"<<endl;
        }
        else
        {
            cout<<"NO"<<endl;
        }
    }
    return 0;

}

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