pat 1136 A Delayed Palindrome

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.



思路：因为他的数字个数不超过1000个，所以用字符串来保存数据；
然后判断是不是回文串，反过来比较是否相同即可；


其中用到了reverse（）函数，把一个字符串反过来，大数相加，存进位；因为是加法，所以进位最大也只可能是1；

注意事项：
将单个字符存入字符串：ans += char(num + '0');
将个位数转换为char类型，要 + '0'；不能 - ‘0’


代码如下：

#include<bits/stdc++.h>
using namespace std;
#define ll long long

string add(string a)
{
    string ans;
    int l = a.length();
    int co = 0;
    for(int i = 0;i < l;i++)
    {
        int num = (a[i] - '0') + (a[l - 1 - i] - '0') + co;
        //cout << num << "---" << endl;
        co = 0;
        if(num > 9)
        {
            co = 1;
            num -= 10;
        }
        ans += char(num + '0');
    }
    if(co == 1)
    ans += '1';
    reverse(ans.begin(),ans.end());
    //cout << ans << endl;
    return ans;
}


string a,b;
int main()
{
    cin >> a;
    int i,len;
    for(i = 0;i < 10;i++)
    {
        b = a;
        reverse(b.begin(),b.end());
        if(a == b)
        break;
        cout << a << " + " << b << " = " << add(a) << endl;
        a = add(a);
    }
    if(i == 10)
    cout << "Not found in 10 iterations." << endl;
    else
    cout << a << " is a palindromic number." << endl;
    return 0;
}