Light OJ -- 1028(circle_in_square)

Think about the prime factor of an number;

9=3,9
10=2,5,10
So, if number is 9 is eligible for 3,9
9 is base 2=1001
9 is base 3=100
9 is base 4=21
9 is base 5=14
9 is base 6=13
9 is base 7=12
9 is base 8=11
9 is base 9=10
continued....
But not gain any zero at last position
9 is base 10=9
///Bismillahir Rahmanir Rahim
///Author:Tanvir Ahmmad
///CSE,Islamic University,Bangladesh
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string>
#include<cstring>
#include<sstream>
#include<cmath>
#include<cstring>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<vector>
#include<iterator>
#include <functional> ///sort(arr,arr+n,greater<int>()) for decrement of array
/*every external angle sum=360 degree
  angle find using polygon hand(n) ((n-2)*180)/n*/
///Floor[Log(b)  N] + 1 = the number of digits when any number is represented in base b
using namespace std;
typedef long long ll;
typedef long long unsigned llu;
#define M 1000010
ll vis[M+2]= {0},arr[M+2],k=0;
void prime()
{
    arr[k++]=2;
    for(ll i=3; i<M; i+=2)
        if(vis[i]==0)
            for(ll j=i*i; j<=M; j+=i+i)
                vis[j]=1;
    for(ll i=3; i<=M; i+=2)
        if(vis[i]==0)
            arr[k++]=i;
}
int main()
{
    prime();
    ll tst;
    ll num;
    scanf("%lld",&tst);
    for(ll j=1; j<=tst; j++)
    {
        ll sum=1;
        scanf("%lld",&num);
        for(ll i=0; i<=M && arr[i]*arr[i]<=num; i++)
        {
            if(num%arr[i]==0)
            {
                ll cnt=0;
                while(num%arr[i]==0)
                {
                    num/=arr[i];
                    cnt++;
                    if(num<=1) break;
                }
                sum*=(cnt+1);
            }
        }
        if(num!=1) sum*=2;
        printf("Case %lld: %lld\n",j,sum-1);
    }
    return 0;
}
///Alhamdulillah 

Problem Link:https://lightoj.com/problem/trailing-zeroes-i