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 nn and k (1≤n≤109,1≤k≤2⋅105).

Output

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

Otherwise, print YES, and then print k positive integers b1​,b2​,…,bk​ such that each of b_ibi​ is a power of two, and ni=1∑k​bi​=n. If there are multiple answers, you may print any of them.

Sample 1

Inputcopy	Outputcopy
9 4	YES 1 2 2 4

Sample 2

Inputcopy	Outputcopy
8 1	YES 8

Sample 3

Inputcopy	Outputcopy
5 1	NO

Sample 4

Inputcopy	Outputcopy
3 7	NO

 

#include
#include
#include
using namespace std;

const int maxn = 2e5 + 5;  // 最大数组大小

int n, k;  // 输入的整数n和k

int main() {
    cin >> n >> k;  // 输入n和k

    if (n < k) {  // 如果n小于k，无法满足要求
        cout << "NO" << endl;
        return 0;
    }

    vector vec(k, 1);  // 创建一个大小为k的vector，初始值都为1
    n -= k;  //保证至少又k个数字可用于构造

    for (int i = 0; i < k; ++i) {
        while (vec[i] <= n) {  // 当前位置的数字还可以翻倍时，继续翻倍直到不满足条件
            n -= vec[i];  // 减去已经使用的数字数量
            vec[i] *= 2;  // 将当前位置的数字翻倍
        }
    }

    if (n) {  // 如果仍然有剩余的数字没有使用完，则无法满足要求
        cout << "NO" << endl;
    }
    else {  // 否则可以满足要求
        cout << "YES" << endl;
        for (int i = 0; i < k; ++i) {
            cout << vec[i] << (i == k - 1 ? '\n' : ' ');  // 输出结果，最后一个数字后面换行，其他数字后面加空格
        }
    }

    return 0;
}