1136 A Delayed Palindrome (20 分) 字符串相加

墨蓝 2024-04-17 05:47 68阅读 0赞

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][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.

**题意：**字符串+字符串反转=新字符串

问能否10次内，新字符串为回文。

#include<stdio.h>
    #include<string>
    #include<string.h>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    int main()
    {
    	int i,j,k;
    	string a,b,c;
    	cin>>a;
    	b=a;
    	reverse(a.begin(),a.end());
    	for(i=0;i<10;i++)
    	{
    		if(a==b)
    		{
    			cout<<a<<" is a palindromic number.\n";
    			break;
    		}
    		c=""; 
    		k=0;
    		for(j=a.size()-1;j>=0;j--)
    		{
    			c+=((a[j]-'0')+(b[j]-'0')+k)%10+'0'; 
    			k=((a[j]-'0')+(b[j]-'0')+k)/10;
    		}
    		if(k) c+='1';//进位
    		reverse(c.begin(),c.end());
    		
    		cout<<b<<" + "<<a<<" = "<<c<<endl;
    		a=c;
    		b=a; 
    		reverse(a.begin(),a.end());
    	}	
    	if(a!=b)
    	cout<<"Not found in 10 iterations.\n";
    }

[https_en.wikipedia.org_wiki_Palindromic_number]: https://en.wikipedia.org/wiki/Palindromic_number