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;
}