对决___

// Problem: 对决
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/11223/E
// Memory Limit: 524288 MB
// Time Limit: 2000 ms
// 2022-03-25 22:29:41
// 
// Powered by CP Editor (https://cpeditor.org)

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

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;   
typedef vector<ll> Vll;               
typedef vector<pair<int, int> > vpii;
typedef vector<pair<ll, ll> > vpll;                        

const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const double pi  = acos(-1.0);

inline ll qmi (ll a, ll b) {
    ll ans = 1;
    while (b) {
        if (b & 1) ans = ans * a%mod;
        a = a * a %mod;
        b >>= 1;
    }
    return ans;
}
inline int read () {
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return f?-x:x;
}
template<typename T> void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) print(x/10);
    putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
    return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
    return (a + b) %mod;
}
inline ll inv (ll a) {
    return qmi(a, mod - 2);
}
const int N = 1e5 +10;
int n, k;
pii a[N];
bool ans[N];

int find (int tar, int l, int r) {
    while (l < r) {
        int mid = l+r +1>>1;
        if (a[mid].fi <= tar) l = mid;
        else r = mid - 1;
    }
    return l;
}
void solve() {
    cin >> n >> k;
    for (int i = 1; i <= n; i ++) {
        cin >> a[i].fi;
        a[i].se = i;
    }
    sort(a + 1, a + 1 + n, [&](pii a, pii b) {
        return a.fi < b.fi;
    });
    
    for (int i = 1; i <= n; i ++) {
        // int j = i;
        // while (j + 1 <= n && a[j + 1].fi == a[j].fi)
            // j ++;
        int j = find (a[i].fi, i, n);
        if (j - 1 >= k) {
            if (a[i].fi * 2 >= a[n].fi / 2)
                ans[a[i].se] = 1;       
        }   
        else {
            int r = k - (j - 1);    
            if (r > 2) continue;
            if (r == 1) {
                if (a[i].fi * 2 >= a[n].fi / 2  || a[i].fi * 4 + 2==a[n].fi)
                    ans[a[i].se] = 1;
            }
            else if (r == 2) {
                if (a[i].fi *2 >= a[n - 1].fi)
                    if (a[i].fi >= a[n].fi / 2)
                        ans[a[i].se] = 1;
                if (a[i].fi  >= a[n - 1].fi / 2)
                    if (a[i].fi * 2 >= a[n].fi)
                        ans[a[i].se] =1;
            }
        }     
    }
    for (int i = 1; i <= n ;i ++) {
        cout << ans[i] << ' ';
    }
    puts("");
}
int main () {
    // ios::sync_with_stdio(0),cin.tie(0), cout.tie(0);
    int t;
    t =1;
    //cin >> t;
    while (t --) solve();
    return 0;
}      

解法;
对于已经比够了k场的
还剩r场的
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