PAT Have fun with numbers (Python)

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本文探讨了如何验证一个特定的九位数在加倍后是否能保持数字的唯一性，并进一步检查是否存在更多具有类似特性的数字。

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 file 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

Analysis:

输入太长，不考虑使用decimal的情况下，求出输入的二倍值

输入和值必须包含相同的元素，每个元素比较拥有相同的个数

Solution:

BTW，回文在python中的判断方法，easy到cry，return s==s[::-1]

def funNum(s):
	dic,dic1,carry,st={},{},0,""
	for i in s[::-1]:
		tmp=int(i)*2+carry
		j=str(tmp%10)
		carry=tmp/10
		st+=j
		if i not in dic:
			dic[i]=1
		else:
			dic[i]+=1
		if j not in dic1:
			dic1[j]=1
		else:
			dic1[j]+=1
	if carry or dic!=dic1:
		print "No"
            	if carry!=0:
            		st+=str(carry)
	else:
		print "Yes"
	print st[::-1]

s=raw_input()
funNum(s)