1024. Palindromic Number (25)

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本文详细介绍了如何通过一系列操作将非回文数转化为回文数，并阐述了实现过程中涉及的关键步骤与算法逻辑。文章还提供了一个实例输入输出分析，帮助读者理解算法的具体应用。

A number that will be the same when it is written forwards or backwards is known as a Palindromic Number. For example, 1234321 is a palindromic number. All single digit numbers are palindromic numbers.

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. For example, if we start from 67, we can obtain a palindromic number in 2 steps: 67 + 76 = 143, and 143 + 341 = 484.

Given any positive integer N, you are supposed to find its paired palindromic number and the number of steps taken to find it.

Input Specification:

Each input file contains one test case. Each case consists of two positive numbers N and K, where N (<= 10¹⁰) is the initial numer and K (<= 100) is the maximum number of steps. The numbers are separated by a space.

Output Specification:

For each test case, output two numbers, one in each line. The first number is the paired palindromic number of N, and the second number is the number of steps taken to find the palindromic number. If the palindromic number is not found after K steps, just output the number obtained at the Kth step and K instead.

Sample Input 1:

67 3

Sample Output 1:

484
2

Sample Input 2:

69 3

Sample Output 2:

13533

注意点：大数用string类型来完成计算

参考代码：

#include <iostream>
#include <string>
#include <stack>
using namespace std;
string N;
int K;
bool isPal(string s){   //whether it is a palindromic
    stack<char> st;
    for(int i=0;i<s.length();i++){
        st.push(s[i]);
    }
    string temp = "";
    while(!st.empty()){
        char ch = st.top();
        st.pop();
        temp = temp+ch;
    }
    return temp==s;
}
string doOperation(string s){
    stack<char> st;
    for(int i=0;i<s.length();i++){
        st.push(s[i]);
    }
    string reverseStr = "";
    while(!st.empty()){
        char ch = st.top();
        st.pop();
        reverseStr = reverseStr+ch;
    }
    int temp[s.length()];
    int flag = 0;
    for(int i=s.length()-1;i>=0;i--){
        int value = s[i]-'0'+reverseStr[i]-'0';
        if(flag==0){    //first to judge whether it is >10
            if(value>9){
                temp[i] = value%10;
                flag =1;
            }else{
                temp[i] = value%10;
                flag = 0;
            }
        }else if(flag==1){
            if(value>8){
                temp[i] = (value+1)%10;
                flag = 1;
            }else{
                temp[i] = (value+1)%10;
                flag = 0;
            }
        }
    }

    string returnValue = "";
    for(int i=0;i<s.length();i++){
        char ch = '0'+temp[i];
        returnValue =returnValue+ch;
    }
    if(flag==1)
        returnValue = "1"+returnValue;
    return returnValue;
}
int main()
{
    cin>>N>>K;
    int i;
    for(i=0;i<K;i++){
        if(isPal(N)){
            cout<<N<<endl<<i;
            break;
        }else{
            N = doOperation(N);
        }
        if(i==K-1){
            cout<<N<<endl<<K;
        }
    }
    return 0;
}