思路：

在1, 2, 3, ... , k的基础上贪心构造。

实现：

1 #include 
     
       2 using namespace std; 3 typedef long long ll; 4 int a[100005]; 5 int main() 6 { 7     int n, k; 8     while (cin >> n >> k) 9     {10         if (n < (ll)k * (k + 1) / 2) { cout << "NO" << endl; continue; }11         if (k == 1) { cout << "YES\n" << n << endl; continue; }12         if (k == 2 && n == 4) { cout << "NO" << endl; continue; }13         for (int i = 1; i <= k; i++) a[i] = i;14         ll r = n - k * (k + 1) / 2;15         int d = r / k, t = r % k;16         for (int i = 1; i <= k; i++) a[i] += d;17         while (t && a[k] < a[k - 1] * 2) { a[k]++; t--; }18         int i = k - 1;19         for ( ; i > k - 1 - t; i--) a[i]++;20         if (a[i + 1] > a[i] * 2) cout << "NO" << endl;21         else22         {23             cout << "YES" << endl;24             for (int i = 1; i <= k; i++) cout << a[i] << " ";25             cout << endl;26         }27     }28     return 0;29 }