标签：模拟
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For a given unordered multiset ofnlowercase English letters ("multi" means that a letter may appearmore than once), we treat all letters as strings of length 1, and repeat thefollowing operationn - 1 times:
- Remove any two elementss and t from the set, and add their concatenation s + t to the set.
The cost of such operation is defined to be, wheref(s, c) denotes the number of times characterc appears in strings.
Given a non-negative integerk,construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the wholeprocess is exactlyk. It can be shown that a solution always exists.
Input
The first and only line of input contains anon-negative integerk (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 therequirements, concatenated to be a string.
Note that the printed string doesn't needto be the final concatenated string. It only needs to represent an unorderedmultiset 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 provedto be the minimum cost of the process.
题意：给你一个字符串序列的最小价值，求该字符串本身，反正题意十分不明确
最小价值：定义每个小写字母在字符串中出现x次，那么cost=sum(x*(x+1)/2)
题目中那些玄学的价值计算方法都是浮云，实际上就是反着倒退模拟，
参考代码
#include<bits/stdc++.h>
using namespace std;
int n,k;
int main()
{
cin>>n;
for(int i=1;i<=26;i++)
{
k=1;
while(k*(k+1)/2<=n)k++;
n-=k*(k-1)/2;
for(int j=1;j<=k;j++)printf("%c",'a'+i-1);
}
return 0;
}
