CodeForces

From Y to Y 

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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 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.


Example
Input
12
Output
abababab
Input
3
Output
codeforces
Note
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.

题意：给你一个数字，你要找出符合条件的字符串。

这个字符串的花费必须最小为n ,

花费计算方法为： , where f(s, c) denotes the number of times character c appears in string s.

这句话的意思大概就是合并t串和s串的时候，花费为每一个字母 在s串和t串中出现次数相乘  之后再 依次相加。

#include<stdio.h>
#include<iostream>
using namespace std;
int main()
{
    int n;
    cin>>n;
    if(n==0)
        cout<<"a"<<endl;
    for(char i='a'; i<='z'&&n; i++)
    {
        for(int j=1;; j++)
        {
            cout<<i;
            if(n<(j*(j+1))/2)
            {
                n-=(j*(j-1))/2;
                break;
            }
        }
        if(n==0)
            return 0;
    }
}