Codeforces 849C From Y to Y【思维】

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.

题目大意：

一个字符串的计算价值的方式是：一开始字符串是每个字符一组，然后每次合并两个字符组，合并一个字符组的价值是：

这里S,T是选择的两个字符组，F（s，c）表示的是字符组S中，出现c这个字符的次数。
我们知道计算方式是多重的，我们现在取总值最小的那个方法。
现在给你一个最小价值k，让你构造一个不超过1e5长度的字符串

思路：

①我们知道，如果我们希望一个字符串的总价值最小，我们肯定是要从左到右依次合并的。

②那么我们如果一个字符串中有x个a的话，那么对应这些字符a贡献的价值是：（x*(x-1)）/2；

那么每种字符出现的位子就已经无所谓了。

③那么对应我们发现，输入的值不大，那么对应我们暴力走一下就行，具体参考代码。

Ac代码：

#include<stdio.h>#include<string.h>using namespace std;int main(){    int n;    while(~scanf("%d",&n))    {        int nowsum=0;        for(int i=1;i<=26;i++)        {            int pos=100;            for(int j=1;j<=100;j++)            {                if(nowsum+j*(j-1)/2>n)                {                    pos=j-1;                    break;                }            }            for(int j=1;j<=pos;j++)printf("%c",i+'a'-1);            nowsum+=(pos*(pos-1))/2;            if(nowsum==n)break;        }        printf("\n");    }}