CF Round 431 C. From Y to Y

题目链接：http://codeforces.com/contest/849/problem/C

题目大意：给出一个数字K，现在要求构造一个只包含小写字母的可重集合使得集合的最小值为K。一个集合的值定义为：将所有字母看作长度为1的字符串，每次从中取出两个字符串s，t，连接后将新串st加入原集合，s，t从中删除，这个过程花费为，最后直到整个集合只包含一个字符串，那么总的花费值为每个过程花费值的累加和。

解题思路：构造一个字符串使其最优累计值为K，那么首先会想，关键在于如何计算最有累计值，但是我们如果将整个过程反过来看，即将一个长度为N的字符串S拆成N个字母的字符集，那么我们可以每次从S中取出一个字母，认为他是最后加入到S中的，之后再从中取出一个……这样，对于K=5，我们构造一个长度x满足x*(x - 1) / 2 <= k的最大x的字符序列a，然后将k减去x * (x - 1) / 2，重复同样的过程，但是每次的字母都不一样，知道k为0. 注意K为0的情况。

不过，这个方法虽然可行但是我无法证明，只是想到了这种解法，感觉是对的。

代码：

const int maxn = 1e6 + 5;
int k;

void solve(){
    if(k == 0){
        printf("a");
    }
    int tmk = k, cnt = 0;
    while(tmk > 0){
        int x = (sqrt(1.0 + 8 * tmk) + 1) / 2;
        tmk -= x * (x - 1) / 2;
        while(x--) printf("%c", 'a' + cnt);
        cnt++;
    }
    puts("");
}
int main(){
    scanf("%d", &k);
    solve();
}

题目：

C. From Y to Y

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 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.