【CodeForces

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 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 构造一个长度为n字符串，让其执行n-1次 如题的操作，每次操作的费用为
现在给其最小的花费，构造出这样花费的字符串。

分析： 我们可以先单独考虑重复的单独的字母进行操作的花费特点
a 0
aa 1
aaa 3
aaaa 6
aaaaa 10
……
然后我们可以发现 n个重复的字母其花费为(n-1)*n/2 ;
然后我们再将情况复杂点
aaab 3+0
aaabb 3+1
aaabbb 3+3
aaabbbb 3+6
aaabbbbb 3+10
……
这个时候我们就可以发现，其花费是不影响的，重复的只算重复的那部分就可以了。
每个字母的花费规律为
0 1 3 6 10 15 21 28 ……
现在问题变成了，给定一个k，在可以随便用上述所有元素的情况下， 使其和为k。 因为题目没有标明构造不出的情况，所以题目一定可以构造成，故我们可以贪心的每次找到第一个小于k的值，然后输出相应数量的字符，再将k减去相应的值，不断重复下去。直到k为0 。
代码

#include<bits/stdc++.h>using namespace std;typedef pair<int,int>pii;#define first fi#define second se#define  LL long long#define fread() freopen("in.txt","r",stdin)#define fwrite() freopen("out.txt","w",stdout)#define CLOSE() ios_base::sync_with_stdio(false)const int MAXN = 1e4;const int MAXM = 1e6;const int mod = 1e9+7;const int inf = 0x3f3f3f3f;int arr[MAXN]; int main(){    CLOSE();//  fread();//  fwrite();    for(int i=1;i<MAXN;i++) arr[i]=arr[i-1]+i; // 先打表出来    int k;cin>>k;    char temp='a';    if(k==0) putchar(temp);//特判一下    while(k){        int cnt=lower_bound(arr,arr+MAXN,k)-arr;        if(arr[cnt]>k) cnt--;// 找到第一个小于k的值        k-=arr[cnt];        cnt++;  // 因为序号为0的时候，是1个a。         while(cnt--) putchar(temp);        temp++;// 改变字符    }     return 0;}