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A11086 | From Y to Y

时间限制1s
内存限制256MB
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题目描述

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 ![](/uploads/acgo/image/c5dd801b4af60bce_6151cc70321f.jpeg), where $f(s,c)$ denotes the number of times character $c$ appears in string $s$ .

Given a non-negative integer $k$ , construct any valid non-empty set of no more than $100000$ letters, such that the minimum accumulative cost of the whole process is exactly $k$ . It can be shown that a solution always exists.

输入格式

The first and only line of input contains a non-negative integer $k$ ( $0<=k<=100000$ ) — the required minimum cost.

输出格式

Output a non-empty string of no more than $100000$ 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.

输入输出样例

输入 #1
12
输出 #1
abababab
输入 #2
3
输出 #2
codeforces
C++ 编辑器
输入
输出