题目

Consider a positive integer N written in standard notation with k+1 digits $a_i$ as $a_k \cdots a_1a_0$ with $0≤a_i<10$ for all i and $a_k>0$. Then N is palindromic if and only if $a_i = a_{k-i}$ for all $i$. Zero is written 0 and is also palindromic by definition.

Non-palindromic numbers can be paired with palindromic ones via a series of operations. First, the non-palindromic number is reversed and the result is added to the original number. If the result is not a palindromic number, this is repeated until it gives a palindromic number. Such number is called a delayed palindrome. (Quoted from https://en.wikipedia.org/wiki/Palindromic_number )

Given any positive integer, you are supposed to find its paired palindromic number.

Input Specification:

Each input file contains one test case which gives a positive integer no more than 1000 digits.

Output Specification:

For each test case, print line by line the process of finding the palindromic number. The format of each line is the following:

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A + B = C

where A is the original number, B is the reversed A, and C is their sum. A starts being the input number, and this process ends until C becomes a palindromic number – in this case we print in the last line C is a palindromic number.; or if a palindromic number cannot be found in 10 iterations, print Not found in 10 iterations. instead.

Sample Input 1:

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97152

Sample Output 1:

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97152 + 25179 = 122331
122331 + 133221 = 255552
255552 is a palindromic number.

Sample Input 2:

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196

Sample Output 2:

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196 + 691 = 887
887 + 788 = 1675
1675 + 5761 = 7436
7436 + 6347 = 13783
13783 + 38731 = 52514
52514 + 41525 = 94039
94039 + 93049 = 187088
187088 + 880781 = 1067869
1067869 + 9687601 = 10755470
10755470 + 07455701 = 18211171
Not found in 10 iterations.

题解

思路

• 照着要求反转字符串就行了
• 注意输出要以字符串的形式
• 不然前面的0会被忽略掉

数据结构

• 没啥数据结构

算法

• 开始十次遍历
  • 每次遍历前把字符串反转一下
  • 如果反转后和原字符串是相等的
    • 输出相等的那段文字并break
  • 否则，求它们俩的和
  • 输出
  • 使原数等于刚刚的和的字符串的表示
• 十次都没有则输出没找到。

代码

• 由于使用Python可以ac，因此只放了Python的解。

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a = input()
for i in range(10):
    b = a[::-1]
    if b == a:
        print(a, "is a palindromic number.")
        break
    sums = int(a) + int(b)
    print(a, "+", b, "=", sums)
    a = str(sums)
else:
    print("Not found in 10 iterations.")