Palindromic Number

  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 (≤) is the initial numer and K (≤) 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 Kinstead.

Sample Input 1:

67 3

Sample Output 1:

484
2

Sample Input 2:

69 3

Sample Output 2:

1353
3

题目解析
  本题给出两个数字n与k,n代表未转换数字,k代表最大转换步数,如果n不是回文数字屎就执行将n反转,n + 反转后的数字,若n还不是回文数字则继续执行上述操作,直到n为回文数字或执行次数超过最大转换步数为止。

  本题数据可能超过int范围,既然既要判断回文又要反转数字,那么索性之间用处理大数的方法——用字符串记录数字。

  之后模拟出加法操作与反转操作,判断执行后的数组是否为回文串即可。

AC代码

 #include <bits/stdc++.h>
using namespace std;
const int MAX = 1e3;
struct Number{ //记录数字
int num[MAX]; //存储数字的每一位(为了方便加法运算倒序存储)
int len; //存储数字位数
Number(){ //构造函数
memset(num, , sizeof(num)); //初始化num每一位都为0
len = ; //位数为0
}
};
Number toNumber(string temp){ //字符串转化为Number
Number a;
a.len = temp.size();
for(int i = ; i < a.len; i++) //倒序存储
a.num[i] = temp[a.len - i - ] - '';
return a;
}
Number add(Number a, Number b){ //加法函数
Number c;
int carry = ; //进位
for(int i = ; i < a.len || i < b.len; i++){ //以较长的长度为界限(当然本题两个数长度相等,其实以任意一个为界都可以)
int temp = a.num[i] + b.num[i] + carry; //两个位置相加后加上进位
c.num[c.len++] = temp % ; //记录新数字的对应位
carry = temp / ; //计算新的进位
}
if(carry != ) //若最高位有进位
c.num[c.len++] = carry;
return c;
}
Number reversed(Number a){ //反转函数
Number b;
for(int i = a.len - ; i >= ; i--){
b.num[b.len++] = a.num[i];
}
return b; }
bool judgePalindromic(Number a){ //判断回文
for(int i = ; i < a.len / ; i++){
if(a.num[i] != a.num[a.len - i - ]) //从两端像中间匹配,若失匹返回false
return false;
}
return true;
}
int main()
{
string str;
int k;
cin >> str >> k; //输入数字与最大转化次数
Number n = toNumber(str); //将数字转化为Number型
int cnt = ; //cnt记录当前转化次数
while(cnt < k && !judgePalindromic(n)){ //转化次数超限或成功转化为回文时结束循环
Number temp = reversed(n); //反转
n = add(n, temp); //相加
cnt++; //转化次数加1
}
for(int i = n.len - ; i >= ; i--) //输出转化后的数字
printf("%d", n.num[i]);
printf("\n");
printf("%d\n", cnt); //输出转化次数
return ;
}

PTA (Advanced Level) 1024 Palindromic Number的更多相关文章

  1. PAT (Advanced Level) 1024. Palindromic Number (25)

    手动模拟加法高精度. 注意:如果输入数字的就是回文,这个时候输出0步. #include<iostream> #include<cstring> #include<cma ...

  2. PTA (Advanced Level)1082.Read Number in Chinese

    Given an integer with no more than 9 digits, you are supposed to read it in the traditional Chinese ...

  3. PAT 甲级 1024 Palindromic Number

    1024. Palindromic Number (25) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue A ...

  4. PAT 甲级 1024 Palindromic Number (25 分)(大数加法,考虑这个数一开始是不是回文串)

    1024 Palindromic Number (25 分)   A number that will be the same when it is written forwards or backw ...

  5. PTA (Advanced Level) 1019 General Palindromic Number

    General Palindromic Number A number that will be the same when it is written forwards or backwards i ...

  6. PAT Advanced 1024 Palindromic Number (25) [数学问题-⼤整数相加]

    题目 A number that will be the same when it is written forwards or backwards is known as a Palindromic ...

  7. 1024 Palindromic Number int_string转换 大整数相加

    A number that will be the same when it is written forwards or backwards is known as a Palindromic Nu ...

  8. 1024. Palindromic Number (25)

    A number that will be the same when it is written forwards or backwards is known as a Palindromic Nu ...

  9. PAT 1024 Palindromic Number[难]

    A number that will be the same when it is written forwards or backwards is known as a Palindromic Nu ...

随机推荐

  1. NGUI学习笔记(一)UILabel介绍

    来个前言: 作为一个U3D程序员,自然要写一写U3D相关的内容了.想来想去还是从UI开始搞起,可能这也是最直观同时也最重要的部分之一了.U3D自带的UI系统,也许略坑,也没有太多介绍的价值,那么从今天 ...

  2. OSPF LSA的详解

    LSA类型的配置与查看 1基本配置 R1(config)#NO IP DO LO R1(config)#NO ENAble PAssword R1(config)#LINe COnsole 0 R1( ...

  3. VxWorks 6.9 内核编程指导之读书笔记 -- VxWorks Small-Footprint Configuration

    什么是Small-footprint? Small-footprint常见关键配置? 如何配置Small-footprint? 什么是Small-footprint? Small-footprint配 ...

  4. 两个关于XML解析报错问题小记

    Caused by: org.xml.sax.SAXParseException: The string "--" is not permitted within comments ...

  5. linux kernel的函数与抽象层

    在数学领域,函数是一种关系,这种关系使一个集合里的每一个元素对应到另一个(可能相同的)集合里的唯一元素. 在C语言中函数也有这种联系.自变量影响着因变量. 在linux内核驱动编程经常会有抽象层的概念 ...

  6. 截图工具 Snagit

    相对于其他截图工具方面,Snagit 一个主要特点是: 滚动截图. 另:同样基于手工绘制的形状截图, 有可能截取文本(测试只 windows在窗口内的目录 要么 文件名 实用). 不管是 web页,是 ...

  7. Android -传统蓝牙通信聊天

    概述 Android 传统蓝牙的使用,包括开关蓝牙.搜索设备.蓝牙连接.通信等. 详细 代码下载:http://www.demodashi.com/demo/10676.html 原文地址: Andr ...

  8. HEOI2013 Segment

    传说中的“李超树”. 大意:给你若干线段,试求横坐标x上的最上方一条线段的编号.无则输出零. 解:用线段树维护. 插入的时候保存自己这个区间上可能成为最大值的线段,被抛弃的则看情况下放. 查询时从最底 ...

  9. ant design table column 设置width不生效解决方案

    当td里的内容超出了width的范围时,会出现width不固定,也就是width不生效 解决方案: 设置scroll的width等于所有列宽之和(scroll={{x: 100%}})

  10. HDU4609:3-idiots(FFT)

    Description Input Output Sample Input Sample Output Solution 题意:给你$n$根木棍,问你任选三根能构成三角形的概率是多少. 写挂sb细节心 ...