LightOj 1028 - Trailing Zeroes (I)

       [ ট্রায়ালিং জিরো বলতে আমি যা বুঝলাম, কোন একটা সংখ্যাকে এমন কোন সংখ্যা দিয়ে ভাগ করলে তার ভাগফল জিরো হবে । এবং ওই সংখ্যাটাই হবে তার বেইস । ]
[ A number you will get mod zero to divide another number. Then, that number will be the base of 1st number....]
formula:: 9=3*3; (x+1)*(y+1)=(0+1)*(2+1)=3; where, x and y is the power of  2 and 3. Then the result is = 3-1=2; so, again, x,y,z are represented bt the power of 2,3 and 5.....
/*
    LightOj 1028 - Trailing Zeroes (I)
    Verdict :: Accepted
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
#include <string>
#include <vector>
#include <cmath>
#include <cctype>
#include <sstream>
#include <set>
#include <list>
#include <stack>
#include <queue>
#include <algorithm>
#define sf scanf
#define pf printf
#define sfint scanf ("%d %d",&a,&b)
#define sfl scanf ("%ld %ld",&a,&b)
#define sfll scanf ("%lld %lld",&a,&b)
#define sfd scanf ("%lf %lf",&a,&b)
#define sff scanf ("%f %f",&a,&b)
#define loop(i,n) for(i=0;i<n;i++)
#define LL long long
#define L long
#define nl puts("")
#define MX 1000005
#define N 100
#define MOD 10000000007
#define pb push_back
#define pi acos(-1.0)
#define sz size()
#define gc getchar ()
#define ps push
#define clr clear
#define bn begin()
#define ed end()

using namespace std;

bool flag[MX];
vector<int>v;
int v_len;

void prime()
{
    int i,j,qrt;

    for (i=4;i<=MX;i+=2)
    {
        flag[i]=true;
    }

    v.push_back(2);

    qrt=(int) sqrt(double(MX));

    for(i=3;i<=qrt;i+=2)
    {
        if (!flag[i])
        {
            for(j=i*i;j<=MX;j+=i+i)
            {
                flag[j]=true;
            }
        }
    }

    for(i=3;i<=MX;i+=2)
    {
        if(!flag[i])
        {
            v.push_back(i);
        }
    }

    v_len=v.size();
}

LL primefacto(LL n)
{
    LL i,s,m=1,k;
    for(i=0;i<v_len and v[i]*v[i]<=n;i++)
    {
        k=0;

        if(n%v[i]==0)
        {
            while(n%v[i]==0)
            {
                k+=1;
                n/=v[i];

                if(n==0 or n==1) break;
            }

            s=k+1;
            m*=s;
        }
    }

    if(n != 1) m*=2;

    return (m);
}

int main()
{
    int t,tc,i;
    prime();
    //loop(i,v_len)cout<<v[i] << " ";
    sf("%d",&tc);

    loop(t,tc)
    {
        LL n;
        sf("%lld",&n);

        LL r=primefacto(n)-1;

        pf("Case %d: %lld\n",t+1,r);
    }

    return 0;
}