1136 A Delayed Palindrome (20 分)

1136 A Delayed Palindrome (20 分)

 

Consider a positive integer N written in standard notation with k+1 digits a​i​​ as a​k​​⋯a​1​​a​0​​ with 0 for all iand a​k​​>0. Then N is palindromic if and only if a​i​​=a​k−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.

 

字符串相加，再来个迭代几次就成。

 

#include <bits/stdc++.h>
#define ll long long int
using namespace std;
string s,st;


string add(string s, string ss){
    string st;
    int an = 0;
    for(int i = s.length()-1 ; i >= 0; --i){
        int x = s[i]-'0';
        int y = ss[i]-'0';
        int ans = x+y+an;
        if(ans < 10){
            st += ans+'0';
            an = 0;
        }else{
            st += (ans%10) + '0';
            an = 1;
        }
    }
    if(an) st += '1';
    reverse(st.begin(), st.end());
    return st;
}


bool ishui(string s){
    for(int i = 0; i < s.length(); i++){
        if(s[i] != s[s.length() -i - 1])
            return false;
    }
    return true;
}

int main(){
    cin >> s;
    st = s;
    reverse(st.begin(), st.end());
    int cnt = 10;
    while(!ishui(s) && cnt--){
        string ans = add(s,st);
        cout <<s<<" + "<<st<<" = "<<ans<<endl;
        s = ans;
        st = s;
        reverse(st.begin(), st.end());
    }
    if(cnt == -1){
        cout <<"Not found in 10 iterations."<<endl;
    }else{
        cout << s<<" is a palindromic number."<<endl;
    }
    return 0;
}