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Here is a simple C program to check if a string is a palindrome:

#include <stdio.h>
#include <string.h>

int main() {
    char str[100], rev[100];
    printf("Enter a string: ");
    fgets(str, sizeof(str), stdin);
    str[strcspn(str, "\n")] = 0; // Remove newline character
    strcpy(rev, str);
    strrev(rev); // Reverse the string
    if (strcmp(str, rev) == 0)
        printf("The string is a palindrome.\n");
    else
        printf("The string is not a palindrome.\n");
    return 0;
}

This program takes a string input, reverses it, and then compares the original string with the reversed string to determine if it is a palindrome.

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Repeating: Until you become expert at this I suggest you do this in two stages (using c and d separately). Suppose there are c digits after the decimal place where the digits are non-repeating, after which you get a repeating pattern of a string of d digits. Then the numerator is the old original string including one lot of the repeated digits minus the original string with none of the repeating digits. The denominator is 10c*(10d - 1), which is a string of d 9s followed by c 0s.For example 123.26159159&hellip; There are 2 digits, "26", after the decimal point before the repeats kick in so c = 2, and the repeating string "159" is 3 digits long so d = 3. So the numerator is 12326159 &ndash; 12326 = 12313833 and the denominator is 99900 Therefore the fraction is 12313833/99900.Finally, you should check to see if the fraction can be simplified.


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Until you become expert at this I suggest you do this in two stages (using c and d separately).Suppose there are c digits after the decimal place where the digits are non-repeating, after which you get a repeating pattern of a string of d digits. Then the numerator is the old original string including one lot of the repeated digits minus the original string with none of the repeating digits. The denominator is 10c*(10d - 1), which is a string of d 9s followed by c 0s.For example 123.26159159&hellip; There are 2 digits, "26", after the decimal point before the repeats kick in so c = 2, and the repeating string "159" is 3 digits long so d = 3. So the numerator is 12326159 &ndash; 12326 = 12313833 and the denominator is 99900 Therefore the fraction is 12313833/99900.Check if the fraction can be simplified.


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Until you become expert at this I suggest you do this in two stages (using c and d separately).Suppose there are c digits after the decimal place where the digits are non-repeating, and followed by a repeating pattern of a string of d digits. Then the numerator is the old original string including one lot of the repeated digits minus the original string with none of the repeating digits. The denominator is 10^c*(10^d - 1), which is a string of d 9s followed by c 0s.For example123.26159159&hellip; There are 2 digits, "26", after the decimal point before the repeats kick in so c = 2, and the repeating string "159" is 3 digits long so d = 3.So the numerator is 12326159 &ndash; 12326 = 12313833and the denominator is 99900Therefore the fraction is 12313833/99900.


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The following procedure is for converting a general repeating decimal, that is, one which does not start repeating straight away. Until you become expert at this I suggest you do this in two stages (using c and d separately). Suppose there are c digits after the decimal place where the digits are non-repeating, and then you get a repeating pattern of a string of d digits. Then the numerator is the old original string including one lot of the repeated digits minus the original string with none of the repeating digits. The denominator is 10^c*(10^d - 1), which is a string of d 9s followed by c 0s.Example123.26159159&hellip; There are 2 digits, "26", after the decimal point before the repeats kick in so c = 2, and the repeating string "159" is 3 digits long so d = 3.So the numerator of the fraction is 12326159 &ndash; 12326 = 12313833and the denominator is 99900Therefore the fraction is 12313833/99900.

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