【PAT甲级】1136 A Delayed Palindrome

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🔑中文翻译：延迟的回文数
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1136 A Delayed Palindrome

Consider a positive integer N written in standard notation with k+1 digits $a_i$ as $a_k\cdots a_1{a}_0$ with $0\le a_i<10$ 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:

A + B = C
1

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:

97152
1

Sample Output 1:

97152 + 25179 = 122331
122331 + 133221 = 255552
255552 is a palindromic number.
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Sample Input 2:

196
1

Sample Output 2:

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.
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题意

这道题是 PAT 1024 那道题的改编，几乎一样。同样想要去获得一个回文数，例如 1234321 。

如果给定的数字不是回文数则需要将该数字加上它的逆序数，例如 123 不是回文数，则要加上它的逆序数 321 。然后再判断加上逆序数之后的数是否为回文数，如果还不是就重复上述操作直至是回文数为止。

但是题目有个限制，上述的加法操作最多只能进行十次，如果十次后还不是回文数就不管了，并且输出 Not found in 10 iterations. ，否则输出 i is a palindromic number. ，其中 i 是最终得到的回文数。

思路

1. 这道题给定的数字很大，如果正常加法会爆 int ，需要用到高精度操作，所以一开始我们用字符串类型读入数字，然后再将字符串中的每一位都放进数组当中，注意要从字符转化回数字即 n[i] - '0' 。另外，我们数组中下标为 0 的位置存的是数字的最高位，例如 12345 放到数组中，下标从 0 到 4 存的是从 5 到 1 的数，即倒着放进数组，这样有利于我们后续的高精度操作。
2. 先判断该数字是否为回文数，如果不是就进行上述的加法操作。另外，我们将数字反转利用了 c++ 的语法。
3. 这里套用了高精度的加法模板，其中有些地方可以省略，因为这道题加法的两个数长度是相同的，所以可以不用同时判断 i 小于 a 和 b 的长度。另外，因为 t 是可能会有进位，所以需要判断 t 是否为 0 ,如果不为 0 还需要加到答案数组中。
4. 在操作的过程中需要输出操作结果，格式为 A + B = C 。
5. 根据最后一个数是否是回文数，输出对应的结果。

代码

#include
using namespace std;

//判断是否为回文数
bool check(vector<int> a)
{
    for (int i = 0, j = a.size() - 1; i < j; i++, j--)
        if (a[i] != a[j])
            return false;
    return true;
}

//高精度加法模板
vector<int> add(vector<int> a, vector<int> b)
{
    vector<int> c;
    for (int i = 0, t = 0; i < a.size() || i < b.size() || t; i++)
    {
        if (i < a.size())  t += a[i];
        if (i < b.size())  t += b[i];
        c.push_back(t % 10);
        t /= 10;
    }
    return c;
}

//打印数字
void print(vector<int> a)
{
    for (int i = a.size() - 1; i >= 0; i--)
        cout << a[i];
}

int main()
{
    string n;
    cin >> n;
    vector<int> a;
    for (int i = n.size() - 1; i >= 0; i--)  a.push_back(n[i] - '0');

    for (int i = 0; i < 10; i++)
    {
        if (check(a))    break;	//先判断是否为回文数
        vector<int> b(a.rbegin(), a.rend());	//将数字逆转
        print(a), cout << " + ", print(b), cout << " = ";
        a = add(a, b);	//执行高精度加法操作
        print(a), cout << endl;
    }

    //输出对应结果
    if (!check(a))    puts("Not found in 10 iterations.");
    else    print(a), cout << " is a palindromic number." << endl;

    return 0;
}
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