Repeating Cipher (Codeforces 1095A)
Tags: Implement, implementation

Polycarp loves ciphers. He has invented his own cipher called repeating.Repeating cipher is used for strings. To encrypt the string $$$s=s_{1}s_{2} \dots s_{m}$$$ ($$$1 \le m \le 10$$$), Polycarp uses the following algorithm:  he writes down $$$s_1$$$ ones,  he writes down $$$s_2$$$ twice,  he writes down $$$s_3$$$ three times,  ...  he writes down $$$s_m$$$ $$$m$$$ times. For example, if $$$s$$$="bab" the process is: "b" $$$\to$$$ "baa" $$$\to$$$ "baabbb". So the encrypted $$$s$$$="bab" is "baabbb".Given string $$$t$$$ — the result of encryption of some string $$$s$$$. Your task is to decrypt it, i. e. find the string $$$s$$$.

Input: The first line contains integer $$$n$$$ ($$$1 \le n \le 55$$$) — the length of the encrypted string. The second line of the input contains $$$t$$$ — the result of encryption of some string $$$s$$$. It contains only lowercase Latin letters. The length of $$$t$$$ is exactly $$$n$$$.It is guaranteed that the answer to the test exists.

Output: Print such string $$$s$$$ that after encryption it equals $$$t$$$.


Sample Inputs:
6
baabbb

10
ooopppssss

1
z

Sample Outputs:
bab

oops

z

Solution (Python):
s = int(input())
m = str(input())
p = ''
(i, j) = (0, 0)
while i < len(m):
    p += m[i]
    j += 1
    i += j
print(p)

Solution (Python):
n = int(input())
t = input()
a = 0
originalText = ''
for i in range(0, n):
    a = a + i
    if a < n:
        originalText = originalText + t[a]
print(originalText)

Solution (Python):
number = int(input())
user_input = input()
s = [0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55]
answer = []
for i in range(len(s)):
    while number > s[i]:
        answer.append(user_input[s[i]])
        break
print(*answer, sep='')