題目

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:

11234567899

Sample Output:

1

2Yes

2469135798

題解

思路

建立一個(gè)哈希表，存放每個(gè)字符出現(xiàn)的次數(shù)

將原數(shù)乘以2后，和哈希表每個(gè)字符出現(xiàn)的次數(shù)作對(duì)比

完事兒嗷

數(shù)據(jù)結(jié)構(gòu)

_dict 是一個(gè)哈希表，對(duì)每個(gè)鍵默認(rèn)值是0

鍵是0到9的字符

值是它出現(xiàn)的次數(shù)

算法

對(duì)作為字符串的原數(shù)，按照每一個(gè)字符出現(xiàn)的次數(shù)添加到哈希表

將原數(shù)乘以2,對(duì)每一個(gè)出現(xiàn)的字符，哈希表相應(yīng)減一

如果哈希表不是每一項(xiàng)都為0,那么就說(shuō)明是No

代碼

因?yàn)槭褂肞ython能AC，因此只放了Python的代碼。

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17from collections import defaultdict

_dict = defaultdict(int)

a = input()

for i in a:

_dict[i] += 1

a = str(int(a) * 2)

for i in a:

_dict[i] -= 1

for i in "0123456789":

if _dict[i] != 0:

print("No")

break

else:

print("Yes")

print(a)

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