# 1079 delayed palindromes (20 points)

Posted by Zeradin on Fri, 08 Oct 2021 01:46:20 +0200

1079 delayed palindromes (20 points)

Given a   k+1   Bit positive integer   N. Write   ak​⋯a1​a0​   The form in which all   i   yes   0≤ai​<10   And   ak​>0. N   It is called a palindrome number if and only if for all   i   yes   ai​=ak−i​. Zero is also defined as a palindrome number.

Non palindromes can also be changed into palindromes through a series of operations. First, reverse the number, and then add the reverse number to the number. If the sum is not a palindrome number, repeat the reverse and add operation until a palindrome number appears. If a non palindrome number can change into a palindrome number, it is called the delayed palindrome number. (definition translated from) https://en.wikipedia.org/wiki/Palindromic_number )

Given any positive integer, this question requires you to find the palindrome number.

### Input format:

Input gives a positive integer with no more than 1000 bits in one line.

### Output format:

The process of outputting the number of palindromes for a given integer line by line. The format of each line is as follows

`A + B = C`

among   A   It's the original number, B   yes   A   Number of reversals, C   It's their sum. A   Start with the integer entered. Repeat until   C   It becomes the palindrome number within 10 steps. At this time, it is output in one line   C is a palindromic number.； Or if you can't get the palindrome number in 10 steps, output it in one line at last   Not found in 10 iterations..

### Input example 1:

`97152`

No blank lines at the end

### Output example 1:

```97152 + 25179 = 122331
122331 + 133221 = 255552
255552 is a palindromic number.```

No blank lines at the end

### Input example 2:

`196`

No blank lines at the end

### Output example 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

No blank lines at the end

This problem is similar to the digital black hole. The trouble lies in the 1000 digit number. We can only use the string to solve it, but if the sum of each bit of the string is > = 10, it needs to carry. And there will be a problem when strlen has 0 in the string array!

So I think of a way to make the character with the result of 0 equal to - 2.

I. first, reverse A. I wanted to use strrev, but I can't use c.

```int zhuanhuan(char* A, int n, char* B)
{
int j = 0;
for (int i = strlen(A) - 1; i >= 0; i--)
{
B[j++] = A[i];        //Transfer the elements of A into B
}
}```

2, Enter the huiwen function, and operate as follows:

```int huiwen(char* A, int n, int t)
{
int j = 0;
char B = { 0 }, C = { 0 };
zhuanhuan(A, n, B);              //Call the zhuanhuan function to obtain the inversion of A and store it in B
if (t >= 10)
{
return 0;
}
if (strcmp(A, B) == 0)             Compare before inversion A And after reversal B Whether it's the same, not the same, continue to execute, the same printf Then jump out.
{
printf("%s is a palindromic number.", A);
return 0;
}
for (int i = 0; i < strlen(A); i++)        //For each bit, the sum of A and B is stored in C. if there is 0 + 0, let 0 = - 2; Because if there is 0 for strlen (c), it will be calculated to 0, and the following numbers will not be calculated into the length.
{
C[j] += B[i] - '0' + A[i] - '0';
if (C[j] == 0)
{
C[j] = -2;
}
if(C[j]>=10)
{
C[j] -= 10;
if (C[j] == 0)
{
C[j] = -2;
}
C[j + 1] += 1;
}
j++;
}
for (int k = 0; k < strlen(C); k++)
{
if (C[k] < 0)
{
C[k] = C[k] + 2 + '0';           //Change all elements in C + '0' into characters and numbers.
}
else
{
C[k] += '0';
}
}
if (t < 1)
{
printf("%s + %s = ", A, B);
}
else
{
printf("%s + %s = ", B, A);
}t++;                                     //t is used to count. When it reaches ten times and does not meet the conditions, it prints F, and then ends the program.
for (int l = strlen(C) - 1; l >= 0; l--)
{
printf("%c", C[l]);
}
printf("\n");
return huiwen(C, strlen(C), t);
}```

Finally, attach the whole code:

```#include<stdio.h>
#include<string.h>
int zhuanhuan(char* A, int n, char* B)
{
int j = 0;
for (int i = strlen(A) - 1; i >= 0; i--)
{
B[j++] = A[i];
}
}

int huiwen(char* A, int n, int t)
{
int j = 0;
char B = { 0 }, C = { 0 };
zhuanhuan(A, n, B);
if (t >= 10)
{
return 0;
}
if (strcmp(A, B) == 0)
{
printf("%s is a palindromic number.", A);
return 0;
}
for (int i = 0; i < strlen(A); i++)
{
C[j] += B[i] - '0' + A[i] - '0';
if (C[j] == 0)
{
C[j] = -2;
}
if(C[j]>=10)
{
C[j] -= 10;
if (C[j] == 0)
{
C[j] = -2;
}
C[j + 1] += 1;
}
j++;
}
for (int k = 0; k < strlen(C); k++)
{
if (C[k] < 0)
{
C[k] = C[k] + 2 + '0';
}
else
{
C[k] += '0';
}
}
if (t < 1)
{
printf("%s + %s = ", A, B);
}
else
{
printf("%s + %s = ", B, A);
}t++;
for (int l = strlen(C) - 1; l >= 0; l--)
{
printf("%c", C[l]);
}
printf("\n");
return huiwen(C, strlen(C), t);
}
int main()
{
char N;
int t = 0;
scanf("%s", N);
huiwen(N, strlen(N), t);
}```

Topics: C