2. 程式人生 > >[CF]Codeforces Round #529 (Div. 3) [CF]Codeforces Round #529 (Div. 3)

[CF]Codeforces Round #529 (Div. 3) [CF]Codeforces Round #529 (Div. 3)

阿新 • • 發佈：2018-12-28

[CF]Codeforces Round #529 (Div. 3)

C. Powers Of Two

Description

A positive integer xx is called a power of two if it can be represented as x=2yx=2y, where yy is a non-negative integer. So, the powers of two are 

">1,2,4,8,16,…1,2,4,8,16,….

You are given two positive integers nn and kk. Your task is to represent nn as the sum of exactlykk powers of two.

Input

The only line of the input contains two integers nn and kk (

n≤109">1≤n≤1091≤n≤109, 1≤k≤2⋅1051≤k≤2⋅105).

output

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

Otherwise, print YES, and then print kk positive integers 

,bk">b1,b2,…,bkb1,b2,…,bk such that each of bibi is a power of two, and ∑i=1kbi=n∑i=1kbi=n. If there are multiple answers, you may print any of them.

Examples

Input

9 4

Output

YES
1 2 2 4

Input

5 1

Output

NO

 

正確解法：

題目的意思是說，從2的次冪中找出k個，使他們相加等於n。

我剛開始想的暴搜，從1 1 1 2 2  4 8 等等一個一個找。

我們只要貪心，從大到小就好，找到一個方案就可以。

於是只要滿足 n-  2的次冪  >=k-1 

如果找到有k個，就可以了。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<string>
 4 #include<cstring>
 5 #include<map>
 6 #include<set>
 7 #include<algorithm>
 8 #include<cmath>
 9 #include<cstdlib>
10 using namespace std;
11 int a[40],f[200100],cnt=0;
12 int n, k; 
13 int main()
14 {
15     a[0] = 1;
16     for (int i = 1; i <= 30; i++)
17         a[i] = 2 * a[i - 1];
18     scanf("%d %d",&n,&k);
19     for (int i = 30; i >= 0; i--)
20     {
21         while (n - a[i] >= k - 1 &&n&&k)
22         {
23             n = n - a[i];
24             f[++cnt] = a[i];
25             k--;
26         }
27     }
28     if (n==0)
29     {
30         cout << "YES" << endl;
31         for (int i = cnt; i>1; i--)
32             printf("%d ",f[i]);
33         printf("%d\n",f[1]);
34     }
35     else cout << "NO" << endl;
36     return 0;
37 }
38
View Code

 

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