Which of the following divisors (polynomial) is used by the cyclic redundancy check algorithm CRC16?

int a, b, number1, tmp_no;

printf("Enter any value between 2 and 15\n");

scanf("%d", &number1);

printf("Enter elements\n");

for (a = 0; a < number1; a++)

{

    scanf("%d", &arr1[a]);

}

for (a = 0; a < number1; a++)

{

    for (b = 0; b < (number1 – a – 1); b++)

    {

        if (arr1[b] > arr1[b + 1])

        {

            tmp_no = arr1[b];

            arr1[b] = arr1[b + 1];

            arr1[b + 1] = tmp_no;

        }

    }

}

for (a = 0; a < number1; a++)

{

    printf("%d\n", arr1[a]);

}

}

Which of the given functions will be performed by the following program?

#include <stdio.h>

void main()

{

int arr1[15];

int a, b, number1, tmp_no;

printf("Enter any value between 2 and 15\n");

scanf("%d", &number1);

printf("Enter elements\n");

for (a = 0; a < number1; a++)

{

    scanf("%d", &arr1[a]);

}

for (a = 0; a < number1; a++)

{

    for (b = 0; b < (number1 – a – 1); b++)

    {

        if (arr1[b] > arr1[b + 1])

        {

            tmp_no = arr1[b];

            arr1[b] = arr1[b + 1];

            arr1[b + 1] = tmp_no;

        }

    }

}

for (a = 0; a < number1; a++)

{

    printf("%d\n", arr1[a]);

}

}

{15,25,30,35},

{66,77,88,99}

};

what would be the output of *(*(num+1))?

Given the array:

int num[3][4]=

{

{3,6,9,12},

{15,25,30,35},

{66,77,88,99}

};

what would be the output of *(*(num+1))?

int *pt1, sm=0;

int a, b, p, z, n=9;

z = n;

pt1 = num1;

for (a = 0; a < z; a++)

{

    for (b = 0; b < z; b++)

    {

        if (a == b)

        {

            continue;

        }

        else if (*(pt1 + a) == *(pt1 + b))

        {

            p = b;

            z--;

            while (p < z)

            {

                *(pt1 + p) = *(pt1 + p + 1);

                p++;

            }

            b = 0;

        }

    }

}

for (a = 0; a < z; a++)

{

    sm=sm+num1[a];

}

printf("%d",sm);

return 0;

}

Analyse the following code and choose its correct output from the given options.

#include<stdio.h>

int main()

{

int num1[9] = {3,2,3,4,5,2,3,9,2};

int *pt1, sm=0;

int a, b, p, z, n=9;

z = n;

pt1 = num1;

for (a = 0; a < z; a++)

{

    for (b = 0; b < z; b++)

    {

        if (a == b)

        {

            continue;

        }

        else if (*(pt1 + a) == *(pt1 + b))

        {

            p = b;

            z--;

            while (p < z)

            {

                *(pt1 + p) = *(pt1 + p + 1);

                p++;

            }

            b = 0;

        }

    }

}

for (a = 0; a < z; a++)

{

    sm=sm+num1[a];

}

printf("%d",sm);

return 0;

}