Codeforces 1095C. Powers Of Two——————multiset

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C. Powers Of Two

A positive integer x is called a power of two if it can be represented as $x=2^y$, where y is a non-negative integer. So, the powers of two are 1,2,4,8,16,….

You are given two positive integers n and k. Your task is to represent n as the sum of exactly k powers of two.

Input

The only line of the input contains two integers n and k (1≤n≤1e9, 1≤k≤2⋅1e5).

Output

If it is impossible to represent n as the sum of k powers of two, print NO.

Otherwise, print YES, and then print k positive integers $\ b_1,b_2,…,b_k$ such that each of bi is a power of two, and $∑_{i=1}^{k}b_i=n$. If there are multiple answers, you may print any of them.

Examples

input
9 4
output
YES
1 2 2 4

input
8 1
output
YES
8

input
5 1
output
NO

input
3 7
output
NO

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题目就是说，给你n和k，让你用k个 $\ 2^n\ ,0\le n \le 31$，使得 $∑_{i=1}^{k}b_i=n$,根据样例可以知道， $b_i$可以使重复的。
我们可以知道，对于任何一个正整数，都可以将其写成二进制的加法形式。比如
$9=(1001)_2 = 2^3+2^1 =8+1=9$，
此时k = 2，但是我们可以将 $2^3$分解为 $2^2+2^2$
于是k = 3，此时 $9=2^2+2^2+2^0$，在将大的分解
于是k = 4，此时 $9 = 2^2+2^1+2^1+2^0$,符合题意，输出

$\;multiset\;$是一种某个数可以重复出现的集合，它通常以平衡二叉树完成.

#include<bits/stdc++.h>
// #define SUBMIT
using namespace std;
typedef long long ll;
int main()
{
    #ifdef SUBMIT
        freopen("in.txt","r",stdin);
        freopen("out.txt","w",stdout);
        ll _begin_time = clock();
    #endif

    int n,k;
    multiset<int> s;
    cin>>n>>k;
    int m = n;
    for(int i=0;m;m>>=1,i++)    if(m&1) s.insert(1<<i);
    // for(multiset<int>::iterator it=s.begin();it!=s.end();++it)
    //     cout<<*it<<" ";
    // cout<<endl;
    if(n<k||s.size()>k)  cout<<"NO"<<endl;
    else
    {
        while(s.size()<k)
        {
            int a = *(--s.end()); s.erase(--s.end());
            if(a==1)    break;
            s.insert(a/2);
            s.insert(a/2);
        }
        cout<<"YES"<<endl;
        while(s.size())
        {
            cout<<*s.begin()<<" ";
            s.erase(s.begin());
        }
        cout<<endl;
    }
    #ifdef SUBMIT
        ll _end_time = clock();
        printf("Time: %I64d ms\n",_end_time - _begin_time);
    #endif
    return 0;
}