【PAT甲级 大整数加法】1024 Palindromic Number (25 分)

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

bool isPalindeomic(string S) {
    
     // 判断回文数
	for(int i = 0;i < S.size()/2;++i)
		if(S[i] != S[S.size() - 1 - i])
			return false;
	return true;
}

string add(string a, string b) {
    
     // 大整数加法
	string rst;
	int carry = 0;
	for(int i = a.size() - 1;i >= 0;--i) {
    
     // 从数的低位往高位遍历
		int t = a[i] + b[i]- 2*'0' + carry;
		carry = t / 10; 
		rst.push_back(t % 10 + '0');
	}
	if(carry > 0) rst.push_back(carry + '0');

	reverse(rst.begin(), rst.end()); // 加完的数是反着存储的，所以要reverse
	return rst;
}

int main() {
    
    
	int maxStep;
	string strNum;
	cin >> strNum >> maxStep;
	int step;
    for(step = 0;step < maxStep;++step) {
    
    
		if(isPalindeomic(strNum)) break;
		string num = strNum;
		reverse(begin(strNum), end(strNum));
		string rev_num = strNum;
		strNum = add(num, rev_num);
	}
	cout << strNum << endl;
	step == maxStep ? cout<<maxStep : cout<<step;

	return 0;
}