1136 A Delayed Palindrome (20分)

最新推荐文章于 2021-06-26 20:48:56 发布

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本文介绍了一种通过反复将非回文数与其反转数相加，直至生成回文数的算法。此过程被称为寻找延迟回文数，文章详细解释了算法步骤，并提供了代码示例，演示如何处理正整数，直至找到配对的回文数。

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.

#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;

bool is_palindromic(string original){

    int flag = true;
    if(original == "0"){

        return true;
    }

    for(int i = 0; i < original.length() / 2; i++){
        if(original[i] != original[original.length() -1 - i]){

            flag = false;
            break;
        }
    }
    return flag;
}

string add_num(string original){
    string reversed = original;
    string result = "";
    reverse(reversed.begin(), reversed.end());
    
    int m = 0;
    for(int i = original.length() - 1; i >= 0; i--){
        
        if(int(reversed[i] - '0') + int(original[i] - '0') + m < 10){
            
            result += to_string(int(reversed[i] - '0') + int(original[i] - '0') + m);
        
            m = 0;
        }else
        {
            result += to_string(int(reversed[i] - '0') + int(original[i] - '0') + m - 10);
            
            m = 1;
        }
        

    }

    if(m == 1){
        result += "1";
    }
    // cout << result << endl;
    reverse(result.begin(), result.end());
    printf("%s + %s = %s\n", original.c_str(), reversed.c_str(), result.c_str());
    return result;
}

int main(){

    string num;
    cin >> num;

    if(is_palindromic(num)){

        printf("%s is a palindromic number.\n", num.c_str());
        return 0;
    }
    for(int i = 0; i < 10; i++){

        num = add_num(num);
        if(is_palindromic(num)){

            printf("%s is a palindromic number.\n", num.c_str());
            return 0;
        }
        

    }
    
    printf("Not found in 10 iterations.\n");


    return 0;
}