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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.
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.
1234567899
结尾无空行
Sample Output:
Yes
2469135798
结尾无空行
N = input()
P = str(int(N)*2)
num1 = {}
num2 = {}
num1 = num1.fromkeys((0,1,2,3,4,5,6,7,8,9),0)
num2 = num2.fromkeys((0,1,2,3,4,5,6,7,8,9),0)
for i in N:
num1[int(i)] += 1
for i in P:
num2[int(i)] += 1
flag = 0
for i in range(10):
if num1[i] != num2[i] and flag == 0:
print('No')
flag = 1
if flag == 0:
print('Yes')
print(P)
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