c语言中的回文序列

Question

I need help to write a code in c. I need to find palindrome in a given length of an int in a given sequence. Like this: The length is 3 And the sequence is 3846401 Here it is 464 Or: Length 4 And sequence: 3400003 Here is 0000

I really need some advice here I have tried to do this for hours and could not make it.

after edit: yes i understand all the downvotes, I looked for some advice in general. but this is my code, my problem is that i found away to arrange the input backward and not as the original and that in causing a problem, if the input of k is 2345 than my array is 5432.

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int s;
int k;
int module;
int last;
int y;
int t=1;
int counter=0;
#define N 10
   
int main() {

    printf("Please enter a length:\n");
    scanf("%d",&s);
    printf("Please enter a sequence:\n");
    scanf("%d",&k);
    int x =k;

    while(x>0) {
        x = x/10;
        y++;
    }

    int sequence[N];
    int last_array[N];
    for(int i=0; i<1;i++){
        last_array[i]=-1;
    }

    for(int i=0; i < y; i++){
        module= k%10;
        sequence[i] = module;
        k=k/10;
    }


    for(int i=0; i < y; i++){
        if( counter<s&&sequence[i]==sequence[i+s-t]){
                last_array[counter]=sequence[i];

                t=2+t;
                counter++;
        }

    }


    if(last_array[0]==-1){
        printf("There is no such palindrome.\n");
    }
    else{
        printf("The requested palindrome is: ");
        for(int i=0; i<s; i++){
            printf("%d",last_array[i]);
        }
    }

    return 0;
}

Answer 1

I get the point of the downvoters, but it seems you are pretty lost. Because you haven't posted any code of yours, I won't post any code either, until you do. I will only try to explain how you would approach the task.

In general there are some ways to determine if a subset of an array is a palindrome. One of the most comprehensible ones, in my opinion, is as follows.

Imagine you have an array like this: [3,8,4,6,4,0,1]

If you want to find a palindrome of length 3, what you can do is to divide the input array into smaller subarrays of length 3. In the example above you would have 5 possible subarrays with the length of the desired palindrome:

[3,8,4] [8,4,6] [4,6,4] [6,4,0] [4,0,1]

As you can see the 3rd one is a valid palindrome for this example. Now how do you determine when there is a valid palindrome? Basically you iterate through all the possible subarrays, split the current subarray in half and check if all the digits on the left half are the same as the digits on the right. If one digit on the left does not equal the corresponding digit on the right, you know it can't be a palindrome. If, for every digit, this condition (left = right) is true, then it is a palindrome.

To help you visualise we define the start as "s" and the end as "e":

This is for an odd array, in this case the 3rd subarray of the example.

1st iteration:

 s   e
 |   |
 v   v
[4,6,4] start=4, end=4 <- They are both the same

2nd iteration (start and end fall on the same digit if it is uneven):

  s/e
   |
   v
[4,6,4] start=6, end=6 <- equal again

For an even arbitrary array it would look like this:

1st iteration:

 s         e
 |         |
 v         v
[4,6,1,1,6,4] start=4, end=4 <- equal

2nd iteration:

   s     e
   |     |
   v     v
[4,6,1,1,6,4] start=6, end=6 <- equal

3rd iteration:

     s e
     | |
     v v
[4,6,1,1,6,4] start=1, end=1 <- equal and finish

If at some point they are not equal, again, you know it can't be a palindrome.

Programatically you would have two loops. The outer loop iterates through all the possible subarrays and the second loop compares the elements. If the second loop at some point determines that s and e are not the same, you know it isn't a palindrome and can continue with the next subarray. If the s and e in the second loop are always the same, you know for sure, that this subarray is a palindrome. How to continue further after you find a palindrome, is up to the task. Either you just return the first palindrome subarray you find or you loop (in the outer loop) through all subarrays to determine if there are any others.

I hope it is a bit clearer now. It would be easier if you could clarify on where you are actually stuck, because no one can know what kind of programming experience you have.

Answer 2

u can check if there's a palindrome subset in the set, by the following code in python :

nums = list(map(int, input().split()))

flag = False
for j in range(len(nums)):
    for k in range(j + 2, len(nums)):
        if nums[j] == nums[k]:
            flag = True
            break

if flag == True:
    print("YES")
else:
    print("NO")

Explanation: if the first and last items in the list are the same then it'll be a palindrome.

• For example: if we have a list like that [1,2,2,1]
• j = 0
• k = 0+2, to skip the second number between them
• if nums[j]==nums[k] which first and the last item in list , then it'll be palindrome.