PAT甲级 1136

1136 A Delayed Palindrome （20 分）

Consider a positive integer N written in standard notation with k+1 digits a​i​​ as a​k​​⋯a​1​​a​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:

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:

97152

Sample Output 1:

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

Sample Input 2:

196

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.

题意：判断回文，且最多 迭代10次

#include <iostream>
#include<bits/stdc++.h>
using namespace std;
//对S进行倒置操作
string rev(string s)
{
    reverse(s.begin(), s.end());//algorithm头文件里面的函数reverse(s.begin(), s.end())直接对s进行倒置
    return s;
}
string add(string s1, string s2)
{
    string s = s1;
    int carry = 0;
    for (int i = s1.size() - 1; i >= 0; i--)
    {
        //s2实际上是s1的倒置
        s[i] = (s1[i] - '0' + s2[i] - '0' + carry) % 10 + '0';
        carry = (s1[i] - '0' + s2[i] - '0' + carry) / 10;
    }
    if (carry > 0) s = "1" + s;//1是在最高位上
    return s;
}
int main()
{
    string s, sum;
    int n = 10;
    cin >> s;
    if (s == rev(s))
    {
        cout << s << " is a palindromic number.\n";
        return 0;
    }
    while (n--)
    {
        sum = add(s, rev(s));
        cout << s << " + " << rev(s) << " = " << sum << endl;
        if (sum == rev(sum))
        {
            cout << sum << " is a palindromic number.\n";
            return 0;
        }
        s = sum;
    }
    cout << "Not found in 10 iterations.\n";
    return 0;
}