1136 A Delayed Palindrome (20 分)（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:

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.

生词

英文	解释
Palindrome	回文

分析：

1 将字符串倒置与原字符串比较看是否相等可知s是否为回文串

2 字符串s和它的倒置t相加，只需从头到尾相加然后再倒置（记得要处理最后一个进位carry，如果有进位要在末尾+’1’）

3 倒置可采用algorithm头文件里面的函数reverse(s.begin(), s.end())直接对s进行倒置

原文链接：https://blog.csdn.net/liuchuo/article/details/79064855

题解

#include <bits/stdc++.h>

using namespace std;
string add(string a){
    string b=a,ans;
    reverse(b.begin(),b.end());
    int len=a.length(),carry=0;
    for(int i=0;i<len;i++){
        int num=(a[i]-'0'+b[i]-'0')+carry;
        carry=0;
        if(num>=10){
            carry=1;
            num-=10;
        }
        ans+=char(num+'0');
    }
    if(carry==1) ans+='1';
    reverse(ans.begin(),ans.end());
    return ans;
}
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("1.txt", "r", stdin);
#endif
    string s;
    cin>>s;
    int cnt=0;
    while(cnt<10){
        string t=s;
        reverse(t.begin(),t.end());
        if(t==s){
            cout<<s<<" is a palindromic number.";
            break;
        }else{
            cout << s << " + " << t << " = " << add(s) << endl;
            s=add(s);
            cnt++;
        }
    }
    if(cnt==10) cout<<"Not found in 10 iterations.";
    return 0;
}