题目：
I. Powers Of Two
time limit per test4 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
A positive integer x is called a power of two if it can be represented as x=2y, 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≤109, 1≤k≤2?105).
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 b1,b2,…,bk such that each of bi is a power of two, and ∑i=1kbi=n. If there are multiple answers, you may print any of them.
Examples
inputCopy
9 4
outputCopy
YES
1 2 2 4
inputCopy
8 1
outputCopy
YES
8
inputCopy
5 1
outputCopy
NO
inputCopy
3 7
outputCopy
NO
题意：
给n，k让你用k个二次幂数表示n；
我们知道9用二次幂数表示就是1 8，样例要用4个二次幂数来表示9就是1 2 2 4：
算是一个半水不水的手速题目：
首先要判断：最少要多少个二次幂数来表示n，如果k小于这个最小值，那就很尴尬了，所以要有特判，至于做法：
首先要用二进制表示n：
我们举个例子：19就是10011；（16+2+1）
这个10011放数组存着，然后就是记录1的个数，这个个数就是我们要用的表示出来n要用的最少的二次幂数的个数：
int i=n;int num=2;int time;int num_1=0;for(time=0;i>0;time++){if(i%num){num_1++;all[time]=1;i-=(num/2);}else all[time]=0;num*=2;}int point=time;//二进制的位数//input finished;int goal;if(k<num_1||k>n)printf("NO\n");代码如上，可以看到是这样获取的；
然后就是做法，得到了19变成10011这个数组，我们怎么做下一步？
假设：k为4；
10011如果变成02011会不会等于原来？显然相等，2个8和1个16是相等的，如果需要k个数，但是我只有num_1个1，如何表示呢，只能从高位到低位一直化，相当于什么？大面额的钱给他敞开了来花;上代码：
完整代码：
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <climits>
#include <queue>
#include <stack>
#include <map>
//鬼畜头文件
using namespace std;
#define INF 0x3f3f3f3f
#define ULL unsigned long long
#define LL long long
//鬼畜define
int n,k;
int all[40];
int main()
{scanf("%d %d",&n,&k);int i=n;int num=2;int time;int num_1=0;for(time=0;i>0;time++){if(i%num){num_1++;all[time]=1;i-=(num/2);}else all[time]=0;num*=2;}int point=time;//二进制的位数//input finished;int goal;if(k<num_1||k>n)printf("NO\n");else{printf("YES\n");for(int time1=point-1;time1>=0;time1--){int P=k-num_1;if(P==0){break;}if(all[time1]<=P){num_1+=all[time1];all[time1-1]+=all[time1]*2;all[time1]=0;goal=time1;}else{num_1+=P;all[time1-1]+=P*2;all[time1]-=P;goal=time1;}}int num=1;for(int time=0;time<=goal;time++){for(int time2=0;time2<all[time];time2++){printf("%d ",num);}num*=2;}printf("\n");}}