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.

题解：
大整数加法

#include<bits/stdc++.h>
using namespace std;
const int maxn=100010;
bool isP(string n){
    int len=n.length();
    for(int i=0;i<len;i++){
        if(n[i]!=n[len-1-i]){
            return false;
        }
    }
    return true;
}

string add(string n,string m){
    int lenn=n.length();
    int lenm=n.length();
    reverse(n.begin(),n.end());
    reverse(m.begin(),m.end());
    string t;
    int carry;
    int num;
    int i;
    carry=0;
    for(i=0;i<lenn||i<lenm;i++){
        num=(n[i]-'0')+(m[i]-'0')+carry;
        carry=num/10;
        t+=num%10+'0';
    }
    if(carry>0){
        t+=carry+'0';
    }
    reverse(t.begin(),t.end());
    return t;
}

int main(){
    string n;
    cin>>n;
    int cnt=0;
    string m;
    string t;
    while(!isP(n)){
        m=n;
        reverse(m.begin(),m.end());
        t=add(n,m);
        printf("%s + %s = %s
",n.c_str(),m.c_str(),t.c_str());
        n=t;
        cnt++;
        if(cnt>=10){
            break;
        }
    }
    if(isP(n)){
        printf("%s is a palindromic number.
",n.c_str());
    }
    else{
        printf("Not found in 10 iterations.
");
    }
    return 0;
}