1136 A Delayed Palindrome (20 分)
Consider a positive integer N written in standard notation with k+1 digits ai as ak⋯a1a0 with 0 for all i and ak>0. Then N is palindromic if and only if ai=ak−i for all i. Zero is written 0 and is also palindromic by definition.
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. Such number is called a delayed palindrome. (Quoted from https://en.wikipedia.org/wiki/Palindromic_number )
Given any positive integer, you are supposed to find its paired palindromic number.
Input Specification:
Each input file contains one test case which gives a positive integer no more than 1000 digits.
Output Specification:
For each test case, print line by line the process of finding the palindromic number. The format of each line is the following:
A + B = C
where A is the original number, B is the reversed A, and C is their sum. A starts being the input number, and this process ends until C becomes a palindromic number -- in this case we print in the last line C is a palindromic number.; or if a palindromic number cannot be found in 10 iterations, print Not found in 10 iterations. instead.
Sample Input 1:
97152
Sample Output 1:
97152 + 25179 = 122331
122331 + 133221 = 255552
255552 is a palindromic number.
Sample Input 2:
196
Sample Output 2:
196 + 691 = 887
887 + 788 = 1675
1675 + 5761 = 7436
7436 + 6347 = 13783
13783 + 38731 = 52514
52514 + 41525 = 94039
94039 + 93049 = 187088
187088 + 880781 = 1067869
1067869 + 9687601 = 10755470
10755470 + 07455701 = 18211171
Not found in 10 iterations.
#include<iostream>
#include<algorithm>
using namespace std; bool isPalindromic(string &str){
int len = str.size();
for(int i = ; i < len/; i++){
if(str[i] != str[len - i - ])
return false;
}
return true;
} string add(const string &A,const string &B){
string C;
int len = A.size();
int carry = ;
for(int i = len - ; i >= ; i--){
int temp = A[i] - '' + B[i] - '' + carry;
C += temp % +'';
carry = temp / ;
}
if(carry != ) C += carry + '';
reverse(C.begin(),C.end());
return C;
} int main(){
string A,B,C;
cin >> A;
int cnt = ;
if(isPalindromic(A)){
cout << A << " is a palindromic number.";
return ;
}
while(cnt--){
B = A;
reverse(A.begin(),A.end());
C = add(A,B);
cout << B << " + " << A << " = " << C << endl;
if(isPalindromic(C)){
cout << C << " is a palindromic number.";
return ;
}
A = C;
}
cout <<"Not found in 10 iterations.";
return ;
}
1136 A Delayed Palindrome (20 分)的更多相关文章
- PAT甲级:1136 A Delayed Palindrome (20分)
PAT甲级:1136 A Delayed Palindrome (20分) 题干 Look-and-say sequence is a sequence of integers as the foll ...
- PAT 1136 A Delayed Palindrome
1136 A Delayed Palindrome (20 分) Consider a positive integer N written in standard notation with k ...
- pat 1136 A Delayed Palindrome(20 分)
1136 A Delayed Palindrome(20 分) Consider a positive integer N written in standard notation with k+1 ...
- PAT 1136 A Delayed Palindrome[简单]
1136 A Delayed Palindrome (20 分) Consider a positive integer N written in standard notation with k+1 ...
- 1136 A Delayed Palindrome (20 分)
Consider a positive integer N written in standard notation with k+1 digits ai as ak⋯a1a0 ...
- 1136 A Delayed Palindrome
题意:略. 思路:大整数相加,回文数判断.对首次输入的数也要判断其是否是回文数,故这里用do...while,而不用while. 代码: #include <iostream> #incl ...
- PAT1136:A Delayed Palindrome
1136. A Delayed Palindrome (20) 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 ...
- PAT_A1136#A Delayed Palindrome
Source: PAT_A1136 A Delayed Palindrome (20 分) Description: Consider a positive integer N written in ...
- PAT A1136 A Delayed Palindrome (20 分)——回文,大整数
Consider a positive integer N written in standard notation with k+1 digits ai as ak⋯a1a0 ...
随机推荐
- Docker 学习笔记_安装和使用MongoDB
一.准备 1.宿主机OS:Win10 64 2.虚拟机OS:Ubuntu18.04 3.账号:docker 二.安装 1.搜索MongoDB镜像 ...
- oracle数据库登录
在做以下操作时,要确保你的数据库环境已经正确安装完成.数据库在实际应用中是比较多的,我们测试人员经常会在前台造一些测试数据,在后台数据库进行验证,当然,不局限于此,数据库也可以作为一个专项测试来谈.反 ...
- URAL 1141. RSA Attack(欧拉定理+扩展欧几里得+快速幂模)
题目链接 题意 : 给你n,e,c,并且知道me ≡ c (mod n),而且n = p*q,pq都为素数. 思路 : 这道题的确与题目名字很相符,是个RSA算法,目前地球上最重要的加密算法.RSA算 ...
- Deep Visual-Semantic Alignments for Generating Image Descriptions(深度视觉-语义对应对于生成图像描述)
https://cs.stanford.edu/people/karpathy/deepimagesent/ Abstract We present a model that generates na ...
- HTML5+CSS3从入门到精通 中文pdf版
HTML5+CSS3从入门到精通是通过基础知识+中小实例+综合案例的方式,讲述了用HTML5+ CSS3设计构建网站的必备知识,相对于专业指南.高级程序设计.开发指南同类图书,本书是一本适合快速入手的 ...
- HTML & CSS设计与构建网站 ([美]达科特) PDF原版
HTML & CSS 设计与构建网站采用有别于许多传统编程书籍的新颖编排方式,将使您收到事半功倍的学习效果.每一页都在短小精悍的示例代码的引导下,简明直观.直截了当地阐述一个新主题. < ...
- android 仿QQ气泡聊天界面
1.现在的QQ,微信等一些APP的聊天界面都是气泡聊天界面,左边是接收到的消息,右边是发送的消息, 这个效果其实就是一个ListView在加载它的Item的时候,分别用了不同的布局xml文件. 2.效 ...
- SxsTrace使用的方法
Windows7平台上有一个强大的SxsTrace工具,可以跟踪调试应用程序运行时需要的动态库的版本和路径. SxsTrace使用的方法: 1.首先必须以Administrator用户身份登录 ...
- .netcore项目部署IIS问题
1.一般根据项目安装对应的runtime 下载地址:https://www.microsoft.com/net/download 2.如果出现502错误可以在cmd命令里面运行如下 如果报错,找到对 ...
- MVC区域路由配置