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; }