Light OJ: 1144 – Ray Gun

Problem Link : http://www.lightoj.com/volume_showproblem.php?problem=1144
Solution Idea:

    There are many observations to make in order to get to a working solution.

  • For every lattice point (i, j), the ray that intersects it is unique and it’s identified by the pair , where g is the gcd of i and j.

  • The problem is now reduced to counting the number of irreducible fractions such that a ≤ N and b ≤ M. This is the same as counting for every i between 1 and N, the amount of numbers in the range [1, M] that are coprime with i.

  • Consider a certain number x with prime factors p1, p2. How do we know how many numbers in range [1, M] are coprime with it? That’s equal to M minus the amount of multiples of p1 minus the amount of multiples of p2 plus the amount of multiples of p1 * p2. This is inclusion-exclusion, and in general, if the amount of elements is even, we add, otherwise, we subtract.

  • So now we have a working (but slow) solution: Iterate over every i in the range [1, N] and for every i, factorize it, try out all combinations of primes and then, for every combination that results in a number k, add if the amount of primes is even or subtract if the amount of primes is odd.

  • The previous approach is very slow for two reasons: You’ll be factorizing each number every time and you’ll be doing a lot of repeated work. Every combination of primes you try out at each step will result in a certain number k. A crucial observation is that the higher exponent of that number k will be 1, because we’re trying combinations of different primes. Another crucial observation is that this number k will be seen times in total. Finally, each time we see it, it will contribute by to the final answer (or if the amount of primes is odd).

  • Knowing all this, we can precalculate a lot of stuff and then solve each test case in O(N). We should precalculate the amount of prime factors of every number in the range [1, 106] (this can be done with a simple sieve), and we should cross out numbers that have some prime with an exponent higher than 1 (in other words, multiples of some square). Once we have precalculated all that, we simply iterate from 1 to N and for every number x that we didn’t cross out, we add (or subtract) to our answer.

  • Final observations: We should add 2 to our answer (the two borders). If N = 0, the answer is 1, except M = 0 too, in which case the answer is 0.


  • (This solution idea is from this link )


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SPOJ: SQFREE – Square-free integers

Problem Link : http://www.spoj.com/problems/SQFREE/

Solution Idea:

Basic operation of mobius function. Generate mobius function. Then think about every number whose mobius function value is not zero. If we squre them then something we can get.


#include <bits/stdc++.h>

#define ms(a,b)          memset(a, b, sizeof(a))
#define pb(a)            push_back(a)

#define ss               second
#define sqr(x)           (x)*(x)

#define SZ(a)            (int)a.size()
#define sf(a)            scanf("%d",&a)
#define sfl(a)           scanf("%lld",&a)


#define TEST_CASE(t)     for(int z=1;z<=t;z++)
#define ll long long

using namespace std;

#define mx 10000007

int ara[mx];
vector<ll>num,mobius,prime;
bitset<mx/2>vis;

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

    for(int i=1;i<y;i++)
    {
        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);

}

void precal()
{
    fill(ara,ara+mx,1);
    for(int i=0;prime[i]*prime[i]<mx;i++)
    {
        ll x=prime[i]*prime[i];
        for(int j=x;j<mx;j+=x)
            ara[j]=0;
    }

    for(int i=0;i<SZ(prime);i++)
    {
        int x=prime[i];
        for(int j=x;j<mx;j+=x)
            ara[j]*=-1;
    }

    for(int i=2;i<mx;i++)
    {
        if(ara[i]==0) continue;
        num.pb(i);
        mobius.pb(ara[i]);
    }

}

int main()
{

//    freopen("in.txt","r",stdin);
//	  freopen("out.txt","w",stdout);
    sieve();
    precal();

    int t;
    sf(t);
    TEST_CASE(t)
    {
        ll n;
        sfl(n);

        ll ans=n;

        for(int i=0;i<SZ(num);i++)
        {
            ll x=num[i];
            ll zz=sqr(x);
            if(zz>n) break;
            int y=mobius[i];
            ans+=mobius[i]*(n/zz);
        }

        printf("%lld\n",ans);

    }

    return 0;
}

UVa 11246 – K-Multiple Free set

Problem Link : https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=2203

Solution Idea:


int MFS(int N,int K)
{
int ret=0;
for(int i=1;N;i=-i)
{
ret+=N*i;
N/=K;
}
return ret;
}

So how does this work? Let us start with the full set {1…N}. We need to remove some numbers from this so that it is a K-multiple free set. For this, let us remove every multiple of K from the set. These are the numbers K,2K… and there are N/K of them. Removing them gives us a K-multiple free set. But we have removed some numbers unnecessarily. Since we already removed K, removing K² was unnecessary. Thus we can put back K²,2K²…, which would be N/K² numbers in total. But this ends up putting both K² and K³ into the set and we need to remove all multiples of K³ now. Proceeding in this fashion, it is easy to see that the cardinality of the final set is N – N/K + N/K² – N/K³…

In general, an input size of N=10⁹ in a mathematical problem should give you the idea that neither the time or space complexity of the solution can be O(N) and you have to come up with some sort of a closed form solution.

This solutino idea is from this link.


#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 LINE_PRINT_CASE  printf("Case %d:\n",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));}
/*------------------------------------------------*/


int main()
{

//    freopen("in.txt","r",stdin);
//	  freopen("out.txt","w",stdout);

    int t;
    sf(t);
    TEST_CASE(t)
    {
        ll n,k;
        sffl(n,k);
        if(k==0)
            pf("0\n");
        else
        {
            ll ans=n;
            ll kk=k;
            int cnt=1;
            while(kk<=n)
            {
                if(cnt%2)
                    ans-=(n/kk);
                else
                    ans+=(n/kk);
                kk*=k;
                cnt++;
            }
            pf("%lld\n",ans);
        }
    }

    return 0;
}