B1079 延迟的回文数

题目：https://pintia.cn/problem-sets/994805260223102976/problems/994805261754023936
也是一遍过，不过这题有几个注意点
判断回文数点方法，直接用reverse转换后双等号就行了

代码

#include<algorithm>
#include<string>
#include<cstdio>
#include<iostream>
using namespace std;
string add(string a){
    string b = a, res = "";
    reverse(b.begin(), b.end());
    int carry = 0;
    for(int i = a.size() - 1; i >= 0; i--){
        int temp = (a[i] - '0' + b[i] - '0' + carry) % 10;
        carry = (a[i] - '0' + b[i] - '0' + carry) / 10;
        res += '0' + temp;
    }
    if(carry != 0) res += '0' + carry;
    reverse(res.begin(), res.end());
    return res;
}
bool judge(string a){
    string b = a;
    reverse(b.begin(), b.end());
    if(b == a) return true;
    else return false;
}
int main(){
    string a;
    cin>>a;
    int count = 0;
    if(judge(a)) cout<< a + " is a palindromic number.";
    else {
        while(!judge(a)){
            if(count == 10) {
                cout<<"Not found in 10 iterations.";
                break;
            }
            string rev_a, add_a;
            rev_a = a;
            reverse(rev_a.begin(), rev_a.end());
            add_a = add(a);
            cout<<a + " + " + rev_a + " = " + add_a<<endl;
            if(judge(add_a)){
                cout<< add_a + " is a palindromic number.";
                break;
            }
            a = add_a;
            count++;
        }
    }
    return 0;
}