1136 A Delayed Palindrome (20 point(s))_1136 a delayed palindrome (20point(s))部分正确-优快云博客

本文链接：https://blog.youkuaiyun.com/coderwait/article/details/100185951

本文介绍了一种通过迭代过程将任意正整数转换为回文数的方法，即延迟回文数。通过反转并累加非回文数，直到结果成为回文数，此过程在不超过10次迭代的情况下进行。文章提供了具体的示例输入输出，并详细解释了实现这一算法的C++代码，包括回文判断和基于字符串的加法。

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1136 A Delayed Palindrome (20 point(s))

Consider a positive integer N written in standard notation with k+1 digits ai as ak⋯a1a0 with 0≤ai<10 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:

Sample Output 1:

97152 + 25179 = 122331
122331 + 133221 = 255552
255552 is a palindromic number.

Sample Input 2:

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.

回文串判断、基于字符串的加法。

注意点：

1. 在每一次迭代之前先判断是不是回文串；

2. 字符串的加法注意考虑进位，注意最终加法结果翻转。

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
bool isPal(string s){
	for(int i=0;i<s.length()/2;i++){
		if(s[i]!=s[s.length()-1-i]) return false;
	}
	return true;
}
string add(string a){
	string b = a;
	reverse(a.begin(),a.end());
	int carry=0;
	string ans="";
	for(int i=0;i<a.length();i++){
		int c = a[i]-'0'+b[i]-'0'+carry;
		carry = c/10;
		ans+=char(c%10 +'0');
	}
	if(carry!=0) ans+=char(carry+'0');
	reverse(ans.begin(),ans.end());
	cout<<b<<" + "<<a<<" = "<<ans<<endl;
	return ans;
}
int main(void){
	string s;cin>>s;
	bool flag = false;
	for(int i=0;i<10;i++){
		if(isPal(s)){//先判断是否回文 
			flag = true;
			cout<<s<<" is a palindromic number."<<endl;
			break;
		}
		s = add(s);
	}
	if(!flag) cout<<"Not found in 10 iterations."<<endl;
	return 0;
}