Diane (Codeforces 1554D)
Tags: Greedy, strings

You are given an integer $$$n$$$. Find any string $$$s$$$ of length $$$n$$$ consisting only of English lowercase letters such that each non-empty substring of $$$s$$$ occurs in $$$s$$$ an odd number of times. If there are multiple such strings, output any. It can be shown that such string always exists under the given constraints.A string $$$a$$$ is a substring of a string $$$b$$$ if $$$a$$$ can be obtained from $$$b$$$ by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 500$$$) — the number of test cases.The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$).It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print a single line containing the string $$$s$$$. If there are multiple such strings, output any. It can be shown that such string always exists under the given constraints.

Note: In the first test case, each substring of "abc" occurs exactly once.In the third test case, each substring of "bbcaabbba" occurs an odd number of times. In particular, "b" occurs $$$5$$$ times, "a" and "bb" occur $$$3$$$ times each, and each of the remaining substrings occurs exactly once.

Sample Inputs:
4
3
5
9
19

Sample Outputs:
abc
diane
bbcaabbba
youarethecutestuwuu

Solution (Python):
for testis in range(int(input())):
    n = int(input())
    if n == 1:
        print('a')
    else:
        s = 'a' * (n // 2)
        if n & 1:
            s += 'bc'
        else:
            s += 'b'
        s += 'a' * (n // 2 - 1)
        print(s)

Solution (Python):
for i in range(int(input())):
    n = int(input())
    print('a' * (n // 2), end='b')
    if n % 2 == 1 and n > 1:
        print('c', end='')
    print('a' * (n // 2 - 1))

Solution (Python):
t = int(input())
for i in range(t):
    n = int(input())
    if n == 1:
        print('a')
    else:
        print('a' * (n // 2) + 'b' + ('c' if n % 2 else '') + 'a' * (n // 2 - 1))