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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 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.

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

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.

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：可能说的不清楚，我们取两个字符串，重复的我们把出现次数记录一下，然后相乘

a和b没有重复的，0*0

aba和b有一个重复的 1*1

然后。。为什么最后等于8了我也没想（为什么不是4*4或者1*1？）

反正最后我们加起来等于n就行

解法：

1 构造当然想最容易的 n=12

a a a a这种合并就很简单，0+1+2+3就行

2 我们拿5个a，花费了10，还差2

3 换个字母b，拿两个b b ,还差1

4 再换个字母c，两个c c 搞定

 #include<bits/stdc++.h>

 using namespace std;

 double x[];

 set<double>Se;

 double ans;

 int main(){

     int n;

     cin>>n;

     string s="";

     if(n==){

         cout<<"a"<<endl;

     }else{

         char c='a';

         while(n){

             int sum=;

             int i=;

             for(i=;sum<=n;i++){

                 sum+=i;

             }

             n-=(sum-i+);

             for(int j=;j<i-;j++){

                 s+=c;

             }

             c++;

         }

         cout<<s<<endl;

     }

     return ;

 }