这道题比较简单，而且@qianmo大佬提供的方法和程序也很简单明了了，所以我简单做解释并翻译为python

python

from collections import deque
from typing import Deque
# 采用collections库的deque数据结构，头弹出的操作更为方便

s: Deque[str] = deque(input())
ans: str = ''

# while len(s) > 0:
#     ans += s.popleft()
#     try:
#         s.append(s.popleft())
#     except IndexError:
#         continue
# 防止队列空而返回IndexError，可采用try语句

while True:
    ans += s.popleft()
    if len(s) < 1:
        break
    s.append(s.popleft())
# 这样的循环类似do - while，更为灵活

print(ans)

#include <bits/stdc++.h>
using namespace std;

typedef queue<char> qc;
qc a;

int main()
{
	string s;
	cin >> s;
	for(auto x:s) a.push(x);
	string ans = "";
	while(!a.empty())
	{
		ans+=a.front();
		a.pop();
		a.push(a.front());
		a.pop();
	}
	cout << ans;
	return 0;
}

s=input()

q=[0]*200

head=0

tail=0

for i in range(len(s)):

q[tail]=s[i]

tail+=1

while head<tail:

print(q[head],end="")

head+=1

q[tail]=q[head]

tail+=1

head+=1

s=input()
que=[""]*200
head,tail=0,0
for i in range(len(s)):
    que[tail]=s[i]
    tail+=1
a=""
while tail!=head:
    print(que[head],end="")
    head+=1
    que[tail]=que[head]
    tail+=1
    head+=1

s=input()
que=[""]*100;head,tail=0,0
for i in range(len(s)):
    que[tail]=s[i]
    tail+=1
while head<tail:
    print(que[head],end="")
    head+=1
    if head<tail:
        que[tail]=que[head]
        tail+=1
        head+=1

s=input() q=[0]*200 head=0 tail=0 for i in range(len(s)): q[tail]=s[i] tail+=1 while head<tail: print(q[head],end="") head+=1 q[tail]=q[head] tail+=1 head+=1

#逃课（ n=input() #换行
for i in range(1,len(n)):n=n[:i]+n[1+i:]+n[i] #换行
print(n)