# Codeforces 849 C. From Y to Y （技巧）

## Description

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 $\sum_{c \in {‘a’,’b’,…,’z’}}f(s,c)\times f(t,c)$ , 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 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.

## Input

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

## Output

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.

## Examples input

12


## Examples output

abababab


## 思路

$aaaaaa$ ，我们挑选任意一种合并方式，其结果为 $1+2+3+4+5=15$ ，即 $cnt=len \times (len-1)/2$ 。

## AC 代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+10;
int cnt[maxn];

void init()
{
for(int i=1; i<=10000; i++)
cnt[i]=i*(i-1)/2;
}

void solve(int n)
{
if(!n)
puts("ab");
else
{
int tot=0;
while(n)
{
int i=1;
while(n>=cnt[i])++i;
--i;
n-=cnt[i];
while(i--)
printf("%c",tot+'a');
tot++;
}
putchar('\n');
}
}

int main()
{
init();
int n;
while(cin>>n)
solve(n);
}