notice that the number 123456789 is a 9-digit number consisting exactly the numbers from 1 to 9, with no duplication. double it we will obtain 246913578, which happens to be another 9-digit number consisting exactly the numbers from 1 to 9, only in a different permutation. check to see the result if we double it again! now you are suppose to check if there are more numbers with this property. that is, double a given number with k digits, you are to tell if the resulting number consists of only a permutation of the digits in the original number.

注意，数字123456789是一个由数字1到9组成的9位数，没有重复。将它翻倍，我们将得到246913578，这恰好是另一个由数字1到9组成的9位数，只是排列不同。现在你应该检查如果我们再次将其翻倍的结果是什么！现在你应该检查是否还有更多具有此属性的数字。也就是说，将一个有k位数字翻倍，你需要判断结果数字是否只由原始数字的排列组成。

相关问题

notice that the number 123456789 is a 9-digit number consisting exactly the

数字123456789是一个9位数，正好是由1到9这9个数字组成的。这种数字被称为“连续数”或“全数字数”。它是一个“1到9全数字数”，意思是它包含了1到9这些数字，并且每个数字都只出现了一次。

这种类型的数字在数学中是很有趣的，因为它们在很多数学问题中都有特殊的性质。例如，它们可以被用来构建其他特殊的数字，或者用来解决一些数字逻辑问题。

在这个特定的例子中，我们可以看到123456789这个数字正好包含了1到9这些连续的数字，而且每个数字都只出现了一次。这种数字在数学游戏、数独等解密游戏中经常被使用，因为它们具有独特的特性。

总之，123456789这个9位数的特殊之处在于它是由1到9这些数字恰好组成的，没有多余的数字，也没有缺少的数字。这种数字给人一种整齐、有序的感觉，同时也展现了数字之间的特殊的关联。

Notice that the number 123456789 is a 9-digit number consisting exactly the numbers from 1 to 9, with no duplication. Double it we will obtain 246913578, which happens to be another 9-digit number consisting exactly the numbers from 1 to 9, only in a different permutation. Check to see the result if we double it again! Now you are suppose to check if there are more numbers with this property. That is, double a given number with k digits, you are to tell if the resulting number consists of only a permutation of the digits in the original number. Input Specification: Each input contains one test case. Each case contains one positive integer with no more than 20 digits. Output Specification: For each test case, first print in a line "Yes" if doubling the input number gives a number that consists of only a permutation of the digits in the original number, or "No" if not. Then in the next line, print the doubled number. Sample Input: 1234567899 Sample Output: Yes 2469135798

To solve this problem, we can first check if the number of digits in the given number is even or odd. If it is odd, then we can immediately output "No" and the doubled number. Otherwise, we can count the number of occurrences of each digit in the original number and the doubled number. If they match, then the doubled number consists of only a permutation of the digits in the original number; otherwise, it does not.

Here's the Python code:

num = input().strip()
double_num = str(int(num) * 2)

if len(num) % 2 == 1:
    print("No")
    print(double_num)
else:
    num_count = [0] * 10
    double_count = [0] * 10
    for digit in num:
        num_count[int(digit)] += 1
    for digit in double_num:
        double_count[int(digit)] += 1
    if num_count == double_count:
        print("Yes")
    else:
        print("No")
    print(double_num)

Note that we convert the doubled number to a string so that we can count the occurrences of each digit using a list of counts. We also use the `strip()` method to remove any leading or trailing whitespace in the input.