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 sumof exactly kk powers of two.
Input
The only line of the input contains two integers nn and kk (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 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
8 1
Output
YES
8
Input
5 1
Output
NO
Input
3 7
Output
NO
题意：给定一个数，让你用1，2，4，8等2的倍数表示出来，并且要用指定的k个数，输出一种即可。
思路：他需要的最小的位数为这个数本身转化成2进制数，其中1的数量为最小需要的二进制数，最多需要多少，是n个，因为都可以用1来表示，然后大一位的数代替即可，故可以表示的范围为2进制1的个数到n，这是可以表示出来的，可以从1进行累加到满足，也可以从高位拆分，高位拆分在一般情况下可能更快
下面是代码：
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>using namespace std;long long int a[505];
int m=0;
int tochange(int x) {while(x) {
a[m++]=x%2;
x/=2;
}
}//转成2进制
int main() {int n,k;
scanf("%d%d",&n,&k);
int ans=0;
tochange(n);
for(int t=m-1; t>=0; t--) {
if(a[t]==1) {
ans++;
}
}
if(k<ans||k>n) {
cout<<"NO"<<endl;
}//不满足的情况
else {
cout<<"YES"<<endl;
if(ans==k) {//如果直接等于就不用拆分了
for(int t=0; t<m; t++) {
if(a[t]==1) {
long long int s1=pow(2,t);
printf("%lld ",s1);
}
}
} else {//从高位拆分，高位-1，低位+2，ans+1
int p=m-1;
while(ans!=k) {
while(a[p]>=1) {
a[p]--;
a[p-1]+=2;
ans++;
if(ans==k)
{
break;
}
}
p--;
}//输出
for(int t=m-1; t>=0; t--) {
if(a[t]>=1) {
for(int j=0; j<a[t]; j++) {
long long int s2=pow(2,t);
printf("%lld ",s2);
}
}
}
}
}
return 0;
}
