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 19 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

首先，我们需要将输入的数转换成字符串，并计算出它的长度。接着，我们将该数乘以2，再将结果转换成字符串，并且将它的长度与原始字符串的长度进行比较。如果长度不同，那么乘以2得到的数字肯定不是原始数字的排列。如果长度相同，那么我们需要逐一检查乘以2得到的数字中的每一个数字，确保它们在原始数字中都存在且出现的次数相同。如果满足这个条件，那么乘以2得到的数字就是原始数字的排列，否则就不是。

下面是实现这个算法的代码：