PAT Have fun with numbers (Python)

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)