CF1157D N Problems During K Days

思路：

在1, 2, 3, ... , k的基础上贪心构造。

实现：

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 int a[];
 int main()
 {
     int n, k;
     while (cin >> n >> k)
     {
         if (n < (ll)k * (k + ) / ) { cout << "NO" << endl; continue; }
         if (k == ) { cout << "YES\n" << n << endl; continue; }
         if (k ==  && n == ) { cout << "NO" << endl; continue; }
         for (int i = ; i <= k; i++) a[i] = i;
         ll r = n - k * (k + ) / ;
         int d = r / k, t = r % k;
         for (int i = ; i <= k; i++) a[i] += d;
         while (t && a[k] < a[k - ] * ) { a[k]++; t--; }
         int i = k - ;
         for ( ; i > k -  - t; i--) a[i]++;
         if (a[i + ] > a[i] * ) cout << "NO" << endl;
         else
         {
             cout << "YES" << endl;
             for (int i = ; i <= k; i++) cout << a[i] << " ";
             cout << endl;
         }
     }
     return ;
 }