C. From Y to Y

C. From Y to Y
time limit per test

1 second

memory limit per test

256 megabytes


standard input


standard output

From beginning till end, this message has been waiting to be conveyed.

For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:

  • Remove any two elements s and t from the set, and add their concatenation s + t to the set.

The cost of such operation is defined to be , where f(s, c) denotes the number of times character cappears in string s.

Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.


The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.


Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.

Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.






For the multiset {'a''b''a''b''a''b''a''b'}, one of the ways to complete the process is as follows:

  • {"ab""a""b""a""b""a""b"}, with a cost of 0;
  • {"aba""b""a""b""a""b"}, with a cost of 1;
  • {"abab""a""b""a""b"}, with a cost of 1;
  • {"abab""ab""a""b"}, with a cost of 0;
  • {"abab""aba""b"}, with a cost of 1;
  • {"abab""abab"}, with a cost of 1;
  • {"abababab"}, with a cost of 8.

The total cost is 12, and it can be proved to be the minimum cost of the process.



2017年9月2日 0 / /
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