PAT--1136 A Delayed Palindrome

思路：大数加法+回文判断。但是有几个测试样例没过。。。大佬们帮忙看看哪里错了

#include <iostream>
#include <string>
#include <algorithm>

using namespace std;

bool palid(string str)
{
    int i = 0;
    while(i < str.size()/2)
    {
        if(str[i] != str[str.size()-1-i])
            return false;
        i++;
    }
    return true;
}

string sum(string a, string b)
{
    if(a.size() < b.size())
    {
        string tmp = a;
        a = b;
        b = tmp;
    }
    int leng = a.size() - b.size();
    for(int i = 0; i < leng; i++)
        b = '0'+b;
    bool flag = false;    //进位位
    int tmp;
    string ans;
    for(int i = a.size()-1; i >= 0; i--)
    {
        tmp = a[i]+b[i]-'0'-'0' + flag;
        if(tmp >= 10)
        {
            tmp -= 10;
            flag = true;
        }
        else
            flag = false;
        ans = (char)(tmp+'0') + ans;
    }
    if(flag)
        ans = '1' + ans;
    return ans;
}

int main()
{
    string str, restr, su;
    while(cin >> str)
    {
        int i;
        bool flag = false;
        for(i = 1; i <= 10; i++)
        {
            restr = str;
            reverse(restr.begin(), restr.end());
            su = sum(str, restr);
            cout << str << " + " << restr << " = " << su << endl;
            if(palid(su))
            {
                flag = true;
                break;
            }
            str = su;
        }
        if(flag)
            cout << su << " is a palindromic number." << endl;
        else
            cout << "Not found in 10 iterations." << endl;
    }
    return 0;
}