2022-11-08个人周赛

B. Jamie and Binary Sequence (changed after round)

y最小,字典序最大

先按二进制分解，最大的为2^x。如果全都拆成2^x-1，拆完之后总个数还比k小，就拆，不然就拆最小的

#include <bits/stdc++.h>
using namespace std;
#define IOS ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr)
#define rep(a, b, c) for(int (a)=(b);(a)<=(c);(a)++)
#define per(a, b, c) for(int (a)=(b);(a)>=(c);(a)--)
#define mset(var, val) memset(var,val,sizeof(var))
#define ll long long
#define int ll
#define fi first
#define se second
#define no "No\n"
#define yes "Yes\n"
#define eb emplace_back
#define endl "\n"
#define pii pair<int,int>
#define pll pair<ll,ll>

const int N = 2e5 + 5;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double pi = acos(-1.0);

priority_queue<int>q;
priority_queue<int,vector<int>,greater<int> >p;

map<int,int>mp;

void solve() {
    int n,k,mx=-100;
    cin>>n>>k;
    per(i,62,0){
        if(n&(1ll<<i)){
            q.push(i);
            mp[i]++;
            if(mp[i])    mx=max(mx,i);
        }
    }
    if(q.size()>k){
        cout<<no;
        return;
    }
    int len=k-q.size();
    while(mp[mx]<=len){
        int tx=q.top();
        len=len-mp[mx];
        while(1){
            int x=q.top();
            if(x!=tx) {
                mp[mx-1]=mp[mx-1]+mp[mx]*2;
                mp[mx]=0;
                mx=mx-1;
                break;
            }
            q.pop();
            q.push(x-1);
            q.push(x-1);
        }
    }
    while(!q.empty()){
        p.push(q.top());
        q.pop();
    }
    for (int i = 0; i < len; ++i) {
        int x=p.top();
        p.pop();
        p.push(x-1);
        p.push(x-1);
    }
    while(!p.empty()){
        q.push(p.top());
        p.pop();
    }
    cout<<yes;
    while(!q.empty()){
        cout<<q.top()<<" ";
        q.pop();
    }
    cout<<endl;
}

signed main() {
    IOS;
    int t = 1;
//    cin >> t;
    while (t--) {
        solve();
    }
    return 0;
}