惭愧啊 菜到DIV3 C都又T又WA得
题目大意给出n和k,让你判断n能否由k个2的幂的和组成,并输出。
正确思路:把n转二进制存数组,因为,所以不断从高位向低位转换并判断就好了.
//#include <bits\stdc++.h>
#include <iostream>
#include <queue>
#include <iomanip>
#include <cmath>
#include <set>
#include <algorithm>
using namespace std;
typedef long long ll;
const long double pi = acos(-1);
const ll maxn = 2*1e5+100;
const ll inf = 1e18;
const ll mod = 998244353;
ll n,k,arr[maxn],numone,len;
void two(ll x)
{
while(x)
{
arr[len] = x%2;
if(arr[len] == 1)
numone++;
len++;
x = x/2;
}
len--;
return ;
}
void slove()
{
if(k < numone)
{
cout << "NO" << endl;
return ;
}
else
{
for(ll i = len; i >= 0; i--)
{
while(arr[i] != 0)
{
//cout << i << " " << arr[i] << endl;
if(numone == k)
{
//cout << numone << endl;
cout << "YES" << endl;
ll temp = 1;
for(ll i = 0; i <= len; i++)
{
while(arr[i] != 0)
{
arr[i]--;
cout << temp << " ";
}
temp = temp*2;
}
return ;
}
if(i == 0)
break;
arr[i]--;
numone++;
arr[i-1]+=2;
}
}
}
cout << "NO" << endl;
return ;
}
int main()
{
ios::sync_with_stdio(false);
cin >> n >> k;
two(n);
/*for(ll i = len; i >= 0; i--)
cout << arr[i] << " ";
cout << endl;*/
slove();
return 0;
}