Submission #2559727

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Source Code Expand

#include <bits/stdc++.h>
using namespace std;

#define NDEBUG
#ifdef DEBUG
#include "cout11.h"
#undef NDEBUG
#endif
#include <cassert>

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> ii;
typedef pair<ll,ll> llll;
typedef pair<double,double> dd;

typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<ii> vii;
typedef vector<vector<ii>> vvii;
typedef vector<ll> vll;

#define sz(a)  int((a).size())
#define pb  push_back
#define FOR(var,from,to) for(int var=(from);var<=(to);++var)
#define rep(var,n)  for(int var=0;var<(n);++var)
#define rep1(var,n)  for(int var=1;var<=(n);++var)
#define repC2(vari,varj,n)  for(int vari=0;vari<(n)-1;++vari)for(int varj=vari+1;varj<(n);++varj)
#define ALL(c)  (c).begin(),(c).end()
#define RALL(c)  (c).rbegin(),(c).rend()
#define tr(i,c)  for(auto i=(c).begin(); i!=(c).end(); ++i)
#define found(s,e)  ((s).find(e)!=(s).end())
#define mset(arr,val)  memset(arr,val,sizeof(arr))
#define mid(x,y) ((x)+((y)-(x))/2)
#define IN(x,a,b) ((a)<=(x)&&(x)<=(b))

#include <vector>
using namespace std;

vector<int> primes;
vector<int> smallest_prime_factor;

int sieve(int nmax){
    primes.clear();
    smallest_prime_factor.assign(nmax+1, 0);

    // ここではあえて愚直な書き方
    for (int n=2; n<=nmax; ++n) {
        if (!smallest_prime_factor[n]) {
            primes.push_back(n);
            for (int kn=n; kn<=nmax; kn+=n) {
                if (!smallest_prime_factor[kn]) {  // ←このチェックをせずに上書きを続けると largest_prime_factor[] が出来上がる
                    smallest_prime_factor[kn] = n;
                }
            }
        }
    }
    return primes.size();
}

// xを素因数分解
vector<int> factorize(int x) {
    vector<int> prime_factors;
    while (x > 1) {
        int s = smallest_prime_factor[x];
        if (!s) break;
        if (prime_factors.empty() || prime_factors.back() != s)
            prime_factors.push_back(s);
        x /= s;
    }
    return prime_factors;
}

int cnt[1000001];

int main() {
    mset(cnt, 0);
    sieve(1e5);

    int N,M; cin >> N >> M;
    // vi a(N);
    rep(i,N) {
        int ai; cin >> ai;
        vi f = factorize(ai);
        int w = f.size();
        int P = 1 << w;
        rep1(p, P-1) {
            int x = 1;
            for (int b=0,m=1; b<w; ++b,m<<=1) {
                if (p & m) {
                    x *= f[b];
                }
            }
            ++cnt[x];
        }
        // cerr << ai << f << endl;
    }

    rep1(i,M){
        // fprintf(stderr, "i=%d\n", i);
        if (i == 1) {
            cout << N << endl;
            continue;
        }
        vi f = factorize(i);
        int w = f.size();
        int P = 1 << w;
        int ans = 0;
        rep1(p, P-1){
            int x = 1;
            int pm = (__builtin_popcount(p) % 2) ? 1 : -1;
            for (int b=0,m=1; b<w; ++b,m<<=1) {
                if (p & m) {
                    x *= f[b];
                }
            }
            // fprintf(stderr, " cnt[%d]=%d\n", x, cnt[x]);
            ans += cnt[x] * pm;
        }
        cout << N-ans << endl;
    }

    return 0;
}

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