PAT 1136 A Delayed Palindrome

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 i and 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.

作者: CHEN, Yue

单位: 浙江大学

时间限制: 400 ms

内存限制: 64 MB

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

string reversestr(string s){
    string res = "";
    for(auto x:s){
        res = x+res;
    }
    return res;
}


string strplus(string a,string b){
    string res = "";
    int len = a.size();
    int jinwei = 0;
    for(int i=len-1;i >= 0;i--){
        int numa = a[i] - '0';
        int numb = b[i] - '0';
        int numsum = numa + numb + jinwei;
        jinwei = numsum/10;
        int add = numsum%10;
        res = to_string(add) + res;
    }
    if(jinwei) res = "1"+res;
    return res;
}


int main(){
    string s;
    cin >> s;
    int cnt = 0;

    while(1){
        string t = reversestr(s);
        if(t == s) {
            printf("%s is a palindromic number.", s.c_str());
            break;
        }

        string kkp = strplus(s,t);

        printf("%s + %s = %s\n",s.c_str(),t.c_str(),kkp.c_str());

        s = kkp;

        cnt++;
        if(cnt >= 10){
            printf("Not found in 10 iterations.");
            break;
        }
    }


    return 0;
}

大数相加突然变得好容易啊。。。