A1136. Delayed Palindrome

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.

#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
typedef struct NODE{
    int num[], len;
    NODE(){
        fill(num, num + , );
        len = ;
    }
}bign;
void add(bign &a, bign &b, bign &c){
    c.len = ;
    b.len = ;
    for(int i = a.len - ; i >= ; i--){
        b.num[b.len++] = a.num[i];
    }
    int carry = ;
    int i;
    for(i = ; i < b.len && i < a.len; i++){
        int sum = a.num[i] + b.num[i] + carry;
        carry = sum / ;
        c.num[c.len++] = sum % ;
    }
    while(i < b.len){
        int sum = carry + b.num[i];
        c.num[c.len++] = sum % ;
        carry = sum / ;
    }
    while(i < a.len){
        int sum = carry + a.num[i];
        c.num[c.len++] = sum % ;
        carry = sum / ;
    }
    if(carry != ){
        c.num[c.len++] = carry;
    }
}
int isReverse(bign a){
    for(int i = , j = a.len - ; i <= j; i++, j--){
        if(a.num[i] != a.num[j])
            return ;
    }
    return ;
}
int main(){
    char ss[];
    scanf("%s", ss);
    bign a, b, c;
    for(int i = strlen(ss) - ; i >= ; i--){
        a.num[a.len++] = ss[i] - '';
    }
    int tag = ;
    if(isReverse(a)){
        for(int i = a.len - ; i >= ; i--){
            printf("%d", a.num[i]);
        }
        printf(" is a palindromic number.");
        return ;
    }
    for(int i = ; i < ; i++){
        add(a,b,c);
        for(int k = a.len - ; k >= ; k--){
            printf("%d", a.num[k]);
        }
        printf(" + ");
        for(int k = b.len - ; k >= ; k--){
            printf("%d", b.num[k]);
        }
        printf(" = ");
        for(int k = c.len - ; k >= ; k--){
            printf("%d", c.num[k]);
        }
        printf("\n");
        if(isReverse(c)){
            tag = ;
            for(int j = c.len - ; j >= ; j--){
                printf("%d", c.num[j]);
            }
            printf(" is a palindromic number.");
            break;
        }else{
            for(int k = ; k < c.len; k++){
                a.num[k] = c.num[k];
            }
            a.len = c.len;
        }
    }
    if(tag == ){
        printf("Not found in 10 iterations.\n");
    }
    cin >> ss;
    return ;
}

总结：

1、大整数相加的问题。要注意的是，在a+b做完之后，要注意检查carry是否为0，如果不为0的话，需要再把carry加上。

2、1230的相反是0123而不是123，所以本题相当于两个待加的数字位数都相同，所以可以直接用两个string做加法，比使用大整数模拟要快一些。