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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.
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
1 #include<bits/stdc++.h>
2 using namespace std;
3 double x[1234];
4 set<double>Se;
5 double ans;
6 int main(){
7 int n;
8 cin>>n;
9 string s="";
10 if(n==0){
11 cout<<"a"<<endl;
12 }else{
13 char c=‘a‘;
14 while(n){
15 int sum=0;
16 int i=0;
17 for(i=0;sum<=n;i++){
18 sum+=i;
19 }
20
21 n-=(sum-i+1);
22 for(int j=0;j<i-1;j++){
23 s+=c;
24 }
25 c++;
26 }
27 cout<<s<<endl;
28 }
29 return 0;
30 }