PAT 1024 Palindromic Number [大整数相加]

最新推荐文章于 2022-02-16 09:40:03 发布

原创最新推荐文章于 2022-02-16 09:40:03 发布 · 293 阅读

0 ·

CC 4.0 BY-SA版权

PAT 专栏收录该内容

100 篇文章

订阅专栏

本文介绍了一种算法，用于找到任意正整数N的配对回文数及其所需步骤。通过将非回文数与其反转数相加，重复此过程直至得到回文数，展示了大整数运算的实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

A number that will be the same when it is written forwards or backwards is known as a Palindromic Number. For example, 1234321 is a palindromic number. All single digit numbers are palindromic numbers.

Non-palindromic numbers can be paired with palindromic ones via a series of operations. First, the non-palindromic number is reversed and the result is added to the original number. If the result is not a palindromic number, this is repeated until it gives a palindromic number. For example, if we start from 67, we can obtain a palindromic number in 2 steps: 67 + 76 = 143, and 143 + 341 = 484.

Given any positive integer N, you are supposed to find its paired palindromic number and the number of steps taken to find it.

Input Specification:

Each input file contains one test case. Each case consists of two positive numbers N and K, where N (≤1010) is the initial numer and K (≤100) is the maximum number of steps. The numbers are separated by a space.

Output Specification:

For each test case, output two numbers, one in each line. The first number is the paired palindromic number of N, and the second number is the number of steps taken to find the palindromic number. If the palindromic number is not found after K steps, just output the number obtained at the Kth step and K instead.

Sample Input 1:

67 3

Sample Output 1:

484
2

Sample Input 2:

69 3

Sample Output 2:

1353
3

----------------------------------这是题目和解题的分割线----------------------------------

N (≤10^10) ，K (≤100)，即使是long long也会溢出，所以要用到大整数的相加。

知识点见【大整数运算】大整数的存储 | 高精度加法 | 高精度减法

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>

using namespace std;

struct bigN
{
	int len;
	int b[100];
	//构造函数初始化 
	bigN()
	{
		memset(b,0,sizeof(b));
		len = 0;
	}
}m;

//字符逆序存整型数组，方便从低位到高位计算 
bigN charToint(char s[],int x)
{
	bigN tmp;
	tmp.len = x;
	for(int i=0;i<x;i++)
		tmp.b[x-1-i] = s[i]-'0';
	return tmp;
}

//判断回文 
int ifPNum(bigN z)
{
	for(int i=0;i<z.len/2;i++)
		if(z.b[i]!=z.b[z.len-1-i]) return 0;
	return 1;
}

//逆序 
bigN reBig(bigN a)
{
	for(int i=0;i<a.len/2;i++)
	{
		int x = a.b[i];
		a.b[i] = a.b[a.len-1-i];
		a.b[a.len-1-i] = x;
	}
	return a;
}

bigN sumBigNum(bigN m,bigN n)
{
	bigN c;
	int t = 0,i; //t用来存储进位
	//较长的为界限 
	for(i=0;i<max(m.len,n.len);i++)
	{
		int tmp = m.b[i]+n.b[i]+t; //同位数字相加+进位 
		c.b[c.len++] = tmp%10; //只存个位 
		t = tmp/10; //新的进位 
	}
	//如果还有进位，说明长度要在最大长度上+1 
	if(t!=0)
		c.b[c.len++] = t;
	return c;
}

int main()
{
	int k,i;
	char s[20];
	scanf("%s%d",&s,&k);
	int len = strlen(s),count = 0;
	m = charToint(s,len);
	bigN temp = m;
	//本身是回文则不需要任何操作 
	if(ifPNum(m))
	{
		printf("%s\n%d\n",s,0);
		return 0;
	}
	else
	{
		while(k--)
		{
			//逆序与自身相加的值重新赋给自身 
			temp = sumBigNum(reBig(temp),temp);
			count++;
			//结果为回文可以提前退出循环 
			if(ifPNum(temp)) break;
		}
	}
	//逆序输出 
	for(i=temp.len-1;i>=0;i--)
		printf("%d",temp.b[i]);
	printf("\n%d",count);
}