Light OJ: 1144 – Ray Gun

Problem Link :

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 )

    #include <bits/stdc++.h>
    #include <ext/pb_ds/assoc_container.hpp>
    #include <ext/pb_ds/tree_policy.hpp>
    #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)             cerr<<#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;
    //using namespace __gnu_pbds;
    //typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
    /*----------------------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
    //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 1000006
    ll mobius[mx];
    bool vis[mx];
    ll n,m;
    void precal()
        for(int i=2;i<mx;)
            for(int j=i+i;j<mx;j+=i)
                if(mobius[i]==0) break;
        for(int i=2;i<mx;i++)
                for(int j=i;j<mx;j+=i)
    int main()
    //    freopen("in.txt","r",stdin);
    //	  freopen("out.txt","w",stdout);
        int t;
            if(n>m) swap(n,m);
            ll ans=n*m;
    //        ms(mobius,0);
    //        ms(vis,0);
            for(ll i=2;i<=n;i++)
                if(vis[i]==1) continue;
                ll x=(m/i)*(n/i);
                else if(mobius[i]%2==1)
    //            for(ll j=i;j<=n;j*=i) vis[j]=1;
            if(n) ans++;
            if(m) ans++;
        return 0;