UVa 10139 – Factovisors

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Problem Link : https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=1080

Solution Idea:

Calculate the prime factor of M. Let a prime is P and it’s power is x. so P^x is a prime factor of M. then check that in N! prime factor of P is grater or equal to x.



#include <bits/stdc++.h>

#define pii              pair <int,int>
#define pll              pair <long long,long long>
#define sc               scanf
#define pf               printf
#define Pi               2*acos(0.0)
#define ms(a,b)          memset(a, b, sizeof(a))
#define pb(a)            push_back(a)
#define MP               make_pair
#define db               double
#define ll               long long
#define EPS              10E-10
#define ff               first
#define ss               second
#define sqr(x)           (x)*(x)
#define D(x)             cout<<#x " = "<<(x)<<endl
#define VI               vector <int>
#define DBG              pf("Hi\n")
#define MOD              1000000007
#define CIN              ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define SZ(a)            (int)a.size()
#define sf(a)            scanf("%d",&a)
#define sfl(a)           scanf("%lld",&a)
#define sff(a,b)         scanf("%d %d",&a,&b)
#define sffl(a,b)        scanf("%lld %lld",&a,&b)
#define sfff(a,b,c)      scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c)     scanf("%lld %lld %lld",&a,&b,&c)
#define stlloop(v)       for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n)        for(int i=0;i<n;i++)
#define loop1(i,n)       for(int i=1;i<=n;i++)
#define REP(i,a,b)       for(int i=a;i<b;i++)
#define RREP(i,a,b)      for(int i=a;i>=b;i--)
#define TEST_CASE(t)     for(int z=1;z<=t;z++)
#define PRINT_CASE       printf("Case %d: ",z)
#define CASE_PRINT       cout<<"Case "<<z<<": "
#define all(a)           a.begin(),a.end()
#define intlim           2147483648
#define infinity         (1<<28)
#define ull              unsigned long long
#define gcd(a, b)        __gcd(a, b)
#define lcm(a, b)        ((a)*((b)/gcd(a,b)))

using namespace std;


/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*------------------------------------------------*/

/*-----------------------Bitmask------------------*/
//int Set(int N,int pos){return N=N | (1<<pos);}
//int reset(int N,int pos){return N= N & ~(1<<pos);}
//bool check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/

#define mx 100000

bitset<mx/2> vis;

vector<int>prime;

void sive()
{
    int x=mx/2,y=sqrt(mx)/2;

    for(int i=1; i<y; i++)
    {
        if(vis[i]==0)
        {
            for(int j=i*(i+1)*2; j<x; j+=(2*i)+1)
                vis[j]=1;
        }
    }

    prime.pb(2);
    for(int i=3; i<mx; i+=2)
        if(vis[i/2]==0)
            prime.pb(i);

}

int main()
{
    CIN;
//    freopen("in.txt","r",stdin);
    ///freopen("out.txt","w",stdout);

    sive();

    ll n,m;
    while(cin>>n>>m)
    {
        ll a,b;
        a=n,b=m;
        ll root=sqrt(m)+1;
        bool test=0;
//        if(m==0) test=1;
        for(int i=0;prime[i]<root;i++)
        {
            if(m%prime[i]==0)
            {
                int cnt=0;
                while(m%prime[i]==0)
                    cnt++,m/=prime[i];
                ll temp=n;
                ll sum=0;
                while(temp)
                {
                    sum+=temp/prime[i];
                    temp/=prime[i];
                }
                if(sum<cnt)
                {
                    test=1;
                    break;
                }
                root=sqrt(m)+1;
            }
        }
        if(m>1)
        {
           ll temp=n;
                ll sum=0;
                while(temp)
                {
                    sum+=temp/m;
                    temp/=m;
                }
                if(sum<1)
                {
                    test=1;
                }
        }

        if(test==0)
            cout<<b<<" divides "<<a<<"!"<<endl;
        else
            cout<<b<<" does not divide "<<a<<"!"<<endl;
    }


    return 0;
}