PAT 1024. Palindromic Number

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之前做過一次，部分錯誤。

今天重做，發現是

ret.push_back( (num[i] + other[i] - 96 + carry) % 10 + 48);
		carry = (num[i] + other[i] - 96 + carry) / 10; // carry之前忘記加了
           

在for循環計算carry時，先前的carry忘記加了。

如果按照點選打開連結中，

sum = s[i]+rev[i]-'0'-'0'+carry;
		res.insert(res.begin(),sum%10+'0');
		carry = sum/10;
           

寫成這樣的形式，錯的幾率就小一些了 

代碼：

#include <iostream>
#include <sstream>
#include <list>
#include <deque>
#include <algorithm>

using namespace std;

string number;
int k;

string change_number(string num)
{
	string ret;

	string other = num;
	reverse(other.begin(), other.end());
	int carry = 0; 
	for (size_t i = 0; i < num.size(); ++ i)
	{
		ret.push_back( (num[i] + other[i] - 96 + carry) % 10 + 48);
		carry = (num[i] + other[i] - 96 + carry) / 10; // carry之前忘記加了
	}
	if (carry == 1)
	{
		ret.push_back('1');
	}
	reverse( ret.begin(), ret.end() );	

	return ret;
}

bool palindromic()
{
	string tmp = number;
	reverse(tmp.begin(), tmp.end());
	return tmp == number;
}

int main()
{
	cin >> number >> k;

	int i;
	for (i = 0; i < k; ++ i)
	{
		if ( palindromic() )	
		{
			break;
		}
		number = change_number(number);
//		cout << number << endl;
	}
	cout << number << endl;
	cout << i << endl;

	system("pause");
	return 0;
}