Codeforces Round #431 (Div. 2) C.From Y to Y

题目原文

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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.

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.

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题目大意

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很明显是构造题，输出一个长度不超过10W的字符串。这个字符串通过n-1次操作而来，操作规则就是从你选定的字符串集合里面选2个元素，将它们合并，然后再放回集合，代价就是所选的两个字符串里面公共的字母出现次数相乘的总和，最后求代价总和为k的字符串。

思路

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其实当相同的字母互相连接的时候代价增长是最快的，有n个相同的字符产生的代价是（n-1）*n/2,所以做法就是假定初始字符为‘a’,找到一个最大的i使得（i-1）*i/2<=k,然后打印i个字符并让k=k-(i-1)*i/2,然后‘a’变成‘b’，重复这个过程，直到k=0。

代码

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#include<bits/stdc++.h>
using namespace std;
string s;
char ch;
int a[1005];
void f(int k){
	for (int i=1;i<=k;i++){
		s=s+ch;
	}
	ch++;
	return ;
}
int main(){
	int k;
	for (int i=1;i<=1000;i++){
		a[i]=(i-1)*i/2;
	}
	cin>>k;
	if (k==0)   cout<<"a";
	else{
		s="";
		ch='a';
		while (k!=0){
			int ans=lower_bound(a+1,a+1000+1,k)-a;
			if (a[ans]==k){
				f(ans);   //打cf的时候这里写成了f(a[ans]),导致我这题没过，后来时间到了发现错了改掉秒过，绝了
				break;
			}
			else{
				ans--;
				f(ans);
				k=k-a[ans];
			}
			
		}
		cout<<s;
	}
	return 0;
}