Codeforces Round #431 (Div. 2) C. From Y to Y

题目：

C. From Y to Y

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

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.

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

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.

 

思路：

　　观察样例1很容易发现对于同一种字母，最小花费显然是是n*(n-1)/2，所以可以尝试用几个n*(n-1)/2去构造k。

　　具体怎么证明没多想，用代码测了下发现都是对的，就写了一发。

　

#include <bits/stdc++.h>

using namespace std;

#define MP make_pair
#define PB push_back
typedef long long LL;
typedef pair<int,int> PII;
const double eps=1e-8;
const double pi=acos(-1.0);
const int K=1e6+7;
const int mod=1e9+7;

int k,cnt,ls,num,sum;
int main(void)
{
    cin>>k;
    if(k==0)
    {
        printf("a\n");
        return 0;
    }
    while(k>0)
    {
        ls=num=sum=0;
        while(k>=sum)
            ls=sum,num++,sum+=num;
        k-=ls;
        for(int i=1; i<=num; i++)
            printf("%c",'a'+cnt);
        if(k<=0)break;
        cnt++;
    }
    return 0;
}