[NOIP1999 Popularization Group] Palindrome Number

Title description:

If a number (with a non-zero first digit) reads the same from left to right as from right to left, we call it a palindromic number.

For example: Given a decimal number 56, add 56 to 65 (that is, read 56 from right to left), and get 121$ which is a palindrome.

Another example: for the decimal number 87:

STEP1：87+78=165
STEP2：165+561=726
STEP3：726+627=1353
STEP4：1353+3531=4884

One step here refers to an N$ addition, and the above example used at least 4 steps to get the palindrome number 4884.

Write a program, given an N (2<=10 or N=16) base number M (within 100$ digits), find the minimum number of steps to get the palindrome number. If it is impossible to get the palindrome within 30$ steps (including 30 steps), then output `Impossible!`.

Input format:

Two lines, respectively N, M.

Output format:

If the palindrome can be obtained within 30$ steps, the output format is `STEP=ans`, where ans is the minimum number of steps to obtain the palindrome.

Otherwise output `Impossible!`.

Example :

Sample input #1
10
87

Sample output #1
STEP=4

Ideas:

This question is a question about the use of a simple algorithm, which is equivalent to finding a high-precision palindrome.

because:

Since it is an N-ary number.

Please change --%10 in high precision processing to %n.

Please change --/10 in high precision plus to /n.

Other calculation methods remain unchanged.

First define the variable:

int n, q[1000001], l, w[1000001], ans;
string s;

q is a high-precision array, w is the reversed array of q, l is the length of the high-precision number, n is the base, ans is the number of steps required, and s is the input high-precision string.

High-precision plus code:

void add(int a[], int b[])//高精加 
{
	for(int i = 1; i <= l; i++)
	{
		a[i] += b[i];
		a[i + 1] += a[i] / n;//进位 
		a[i] %= n;
	}
	if(a[l + 1] > 0)//考虑从最高位进位到最高位的下一位 
	{
		l++;//长度++ 
	}
}

High-precision reverse code:

void turn(int a[])//反转数字 
{
	int j = 0;
	for(int i = l; i >= 1; i--)//反着存 
	{ 
		w[++j] = a[i];//存到w数组里 
	}
}

High-precision judgment palindrome code:

bool f(int a[])//判断是否是回文数 
{
	int ln = l;
	int i = 1;//从两边判断
	int j = l;
	while(ln--)
	{
		if(ln < l / 2)//判一般就可以啦QAQ 
		{
			break;
		}
		if(a[i] != a[j])
		{
			return false;//有一位不相等就不是回文数 
		}
		i++;
		j--;
	}
	return true;
}

Array storage high-precision code:

void init()//把s字符串附到q数组里 
{
	int j = 0;
	for(int i = s.length() - 1; i >= 0 ; i--) 
	{
		if(s[i] >= '0' && s[i] <= '9')//数字 
		{
			q[++j] = s[i] - '0';
		}
		else//还有十六进制的 
		{
			q[++j] = s[i] - 'A' + 10;
		} 
	}
}

main main function code:

int main()
{
	cin>>n>>s;
	init();//初始化数组 
	l = s.length();
	while(!f(q))//是否回文 
	{
		turn(q);
		add(q, w);//加上回文数 
		ans++;
		if(ans > 30)//步数大于三十就退出 
		{
			break;
		}
	}
	if(ans > 30)
	{
		printf("Impossible!"); //叹号注意 
	}
	else
	{
		printf("STEP=%d", ans);
	}
	return 0;
}

Full code:

#include <bits/stdc++.h> 
using namespace std;
int n, q[1000001], l, w[1000001], ans;
string s;
void init() 
{
	int j = 0;
	for(int i = s.length() - 1; i >= 0 ; i--) 
	{
		if(s[i] >= '0' && s[i] <= '9') 
		{
			q[++j] = s[i] - '0';
		}
		else 
		{
			q[++j] = s[i] - 'A' + 10;
		} 
	}
}
void add(int a[], int b[]) 
{
	for(int i = 1; i <= l; i++)
	{
		a[i] += b[i];
		a[i + 1] += a[i] / n; 
		a[i] %= n;
	}
	if(a[l + 1] > 0) 
	{
		l++; 
	}
}
bool f(int a[]) 
{
	int ln = l;
	int i = 1;
	int j = l;
	while(ln--)
	{
		if(ln < l / 2) 
		{
			break;
		}
		if(a[i] != a[j])
		{
			return false; 
		}
		i++;
		j--;
	}
	return true;
}
void turn(int a[]) 
{
	int j = 0;
	for(int i = l; i >= 1; i--) 
	{
		w[++j] = a[i]; 
	}
}
int main()
{
	cin>>n>>s;
	init(); 
	l = s.length();
	while(!f(q)) 
	{
		turn(q);
		add(q, w); 
		ans++;
		if(ans > 30) 
		{
			break;
		}
	}
	if(ans > 30)
	{
		printf("Impossible!"); 
	}
	else
	{
		printf("STEP=%d", ans);
	}
	return 0;
}

Summarize:

This question requires many basic algorithms such as high-precision addition, high-precision inversion, high-precision judgment of palindrome numbers, and high-precision array storage.