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
- 932F Escape Through Leaf
传送门 题目大意 https://www.luogu.org/problemnew/show/CF932F 分析 我们可以从叶子向根每次插入b和ans 所以我们不难发现就是相当于插入线段 于是李超树+ ...
- js实现选项卡切换
<!DOCTYPE html><html><head lang="en"> <meta charset="UTF-8" ...
- Python2.X和Python3.X文件对话框、下拉列表的不同
Python2.X和Python3.X文件对话框.下拉列表的不同 今天初次使用Python Tkinter来做了个简单的记事本程序.发现Python2.x和Python3.x的Tkinter模块的好多 ...
- ASP.NET jquery-1.9.1 语句
<script src="Script/jquery-1.9.1.js"></script> <script language="javas ...
- Lucene的基本概念----转载yufenfei的文章
Lucene的基本概念 Lucene是什么? Lucene是一款高性能.可扩展的信息检索工具库.信息检索是指文档搜索.文档内信息搜索或者文档相关的元数据搜索等操作. 信息检索流程如下: 1. 将即将检 ...
- Http报头中不能添加中文字符
今逢一Bug,如下: Invalid non-ASCII or control character in header: 0x6D4B 大意为:报头中有非法字符.故可将其编码后,set入Header, ...
- HBase基准性能测试报告
作者:范欣欣 本次测试主要评估线上HBase的整体性能,量化当前HBase的性能指标,对各种场景下HBase性能表现进行评估,为业务应用提供参考.本篇文章主要介绍此次测试的基本条件,HBase在各种测 ...
- 使用C/C++代码编写Python模块
假如我们要用C语言实现下面的python脚本bird.py import os def fly(name): print(name + " is flying.\n") 调用脚本m ...
- Setter
这个还是比较好理解的. 设置器. 用法还是比较简单的. 语法特征: 设置属性[Property] 填充值[Value] 注意这个是封闭单行闭合标签,可以换行,但只允许在同一个标签闭合. 事例用法: & ...
- windows下安装newman
1.下载安装node.js,下载地址::https://nodejs.org/en/download/,这里我下载的为v10.15.0-x64.msi,下载后直接安装即可,安装完后可输入node -v ...