pat 1136

1136 A Delayed Palindrome (20分)

Consider a positive integer N written in standard notation with k+1 digits ai?? 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:

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.

题意：给定一个数，判断是否为水仙花数，若不是，用这个数和它的逆置数的和取代它继续判断，10次取代之后如果扔不是，结束程序

思路：algorithm数组下的reverse(s.begin(),s.end())函数可以逆置字符串。

代码瑞如下：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<string>
using namespace std;
string rev(string s){
    reverse(s.begin(),s.end());
    return s;
}
string add(string s1,string s2){
    int count=0;
    int len=s1.length();
    string s=s1;
    for(int i=len-1;i>=0;i--){
        s[i]=(s1[i]-'0'+s2[i]-'0'+count)%10+'0';
        count=(s1[i]-'0'+s2[i]-'0'+count)/10;
    }
    if(count==1)
        s="1"+s;
    return s;
}
int main(){
    string s,sum;
    cin>>s;
    if(s==rev(s)){
        cout<<s<<" is a palindromic number." ;
        return 0;
    }
    int mark=0;
    for(int i=0;i<10;i++){
        sum=add(s,rev(s));
        cout<<s<<" + "<<rev(s)<<" = "<<sum<<endl;
        s=sum;
        if(s==rev(s)){
            cout<<s<<" is a palindromic number." ;
            mark=1;
            break;
        }
    }
    if(mark==0){
        cout<<"Not found in 10 iterations.";
    }
    
    return 0;
}