Light OJ 1035 Intelligent Factorial Factorization Solution

   ///La ilaha illellahu muhammadur rasulullah

///******Bismillahir-Rahmanir-Rahim******///

///Abul Hasnat  Tonmoy

///Department of CSE,23rd batch

///Islamic University,Bangladesh

Solution in c++:

#include <bits/stdc++.h>

using namespace std;

#define max 10009

int a[max],ar[max];

int prime[max];

int main()

{

    int t,n,i,k,count=0,j,temp,c=0,v,lastprime;

    for(i=3; i<=sqrt(max); i+=2)

    {

        if(a[i]==0)

        {

          for(j=i*i; j<=max; j+=i)

            {

                a[j]=1;

            }

        }

    }

    prime[count++]=2;

    for(i=3; i<=max; i+=2)

    {

        if(a[i]==0)

        {

            prime[count++]=i;

        }

    }

    cin>>t;

    for(k=1; k<=t; k++)

    {  c=0;

        cin>>n;

        for(i=0; prime[i]<=n; i++)

        {

            temp=n;

            v=0;

            while(temp!=0)

            {

                v+=temp/prime[i];

                temp/=prime[i];

            }

            ar[c]=v;

            c++;

        }

        lastprime=prime[i-1];

        printf("Case %d: %d = ",k,n);

        for(j=0;prime[j]<=n;j++)

        {

            printf("%d (%d)",prime[j],ar[j]);

            if(prime[j]<lastprime)

            {

                printf(" * ");

            }

        }

        cout<<endl;

    }

}