英文:
Flipping The Pairs
问题
以下是翻译好的部分：
程序要求输入一个数字，然后打印出从1到该数字的所有正整数值。但是，数字的顺序会改变，使得每对数字都被翻转。也就是说，2出现在1之前，4出现在3之前，依此类推。例如：
请输入一个数字：6
2
1
4
3
6
5
num = int(input("请输入一个数字："))
r = 2
t = 1
while True:
if t >= num:
break
print(r)
print(t)
r += 2
t += 2
这段代码对偶数运行良好，但在处理奇数时会出现问题。例如，对于输入5，结果应该是：
请输入一个数字：5
2
1
4
3
5
但实际上我的输出是：
请输入一个数字：5
2
1
4
3
希望这可以帮助你。
英文:
The program asks for a number and then prints out all the positive integer values from 1 up to the number. However, the order of the numbers is changed so that each pair or numbers is flipped. That is, 2 comes before 1, 4 before 3 and so forth. For example:
Please type in a number: 6
2
1
4
3
6
5
num = int(input("Please type in a number: "))
r = 2
t = 1
while True:
if t >= num:
break
print(r)
print(t)
r += 2
t += 2
The code works fine with even numbers but then struggles with odd numbers. For example with the input 5 the result should be:
Please type in a number: 5
2
1
4
3
5
But instead my output comes out as:
Please type in a number: 5
2
1
4
3
Any help would be appreciated.
答案1
得分: 1
当num为奇数且t == num时，例如当num为5且t变为5时，循环会在第3次执行之前简单地退出。如果期望的输出是输出5而不是6（因为没有什么需要“翻转”的），你可以在num为奇数时简单地执行print(num)。然而，这与num=6的情况不一致，因为第五次输出不同，所以我建议你仔细检查一下。
英文:
Looking at your own example, when num is odd and t == num, for example when num is 5 and t becomes 5, then the loop simply exits without executing for the 3rd time. If the expected output is to output 5 instead of 6 (because there is nothing to "flip"), you can simply do print(num) if num is odd. However this is inconsistent with the num=6 case as the fifth output is different, so I would recommend you double-check this.
答案2
得分: 0
这里是翻译好的部分：
也许会有几种解决方案，但这里有一个简单的修复方法：
num = 5 # 用于简单测试的赋值
r = 2
t = 1
while t <= num:
if t != num: # 如果t == num，则跳过下面的行
print(r)
print(t)
r += 2
t += 2
一行代码的示例：
print(sorted(range(1, num+1), key=lambda x: x + 2 * (x % 2)))
使用列表推导式：
def fn(x):
return x + 2 * (x % 2) - 1
print([x if x==num and num%2 else fn(x) for x in range(1, num+1)])
以上示例的更可读版本：
output = [x + 2 * (x % 2) - 1 for x in range(1, num+1)]
if num % 2:
output[-1] = num
print(output)
英文:
Maybe there will be several solutions, but a simple fix is here:
num = 5 # assigned for simple testing
r = 2
t = 1
while t <= num:
if t != num: # if t == num, skip the below line
print(r)
print(t)
r += 2
t += 2
An one-liner example:
print(sorted(range(1, num+1), key=lambda x: x + 2 * (x % 2)))
With list comprehension:
def fn(x):
return x + 2 * (x % 2) - 1
print([x if x==num and num%2 else fn(x) for x in range(1, num+1)])
More readable version of the above example:
output = [x + 2 * (x % 2) - 1 for x in range(1, num+1)]
if num % 2:
output[-1] = num
print(output)
答案3
得分: 0
你只能在r <= num时才能打印r，并在其后执行t > num的检查：
num = int(input("请输入一个数字："))
r = 2
t = 1
while True:
if r <= num:
print(r)
if t > num:
break
print(t)
r += 2
t += 2
英文:
You can print(r) only when r <= num, and move the t > num check after it:
num = int(input("Please type in a number: "))
r = 2
t = 1
while True:
if r <= num:
print(r)
if t > num:
break
print(t)
r += 2
t += 2
答案4
得分: 0
你可以对你的现有代码进行简单更改，如下所示：
num = int(input("请输入一个数字："))
r = 2
t = 1
while True:
if t >= num:
if num % 2 != 0:
print(r - 1) # 显示最后一个奇数
break
print(r)
print(t)
r += 2
t += 2
输出将如所期望地显示。
英文:
you can do a simple change to your exiting code like below
num = int(input("Please type in a number: "))
r = 2
t = 1
while True:
if t >= num:
if num % 2 != 0:
print(r-1) # Display the last odd number
break
print(r)
print(t)
r += 2
t += 2
The output will come as desired
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