The output of the following code int a[]={10,20,30,40}, *p; p= a+%3; ++p; printf("%d",*p++);
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![The output of the following code
int a[]={10,20,30,40}, *p;
p = a+1;
++p;
printf("%d",*p++);
Select one:
O a. 10
O b. 30
O C. 40
O d. 20](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3a7af8bb-24eb-482c-ba3b-b6dd32849fd3%2F3b44af71-101a-4c1a-995a-a8001e96960c%2Fmyqvm9e_processed.jpeg&w=3840&q=75)
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- #include <stdio.h>int main(){int d;int s[20],i, j, p, lg=0,m,t=0;char c;printf("Enter number of sale: ");scanf("%d", &d);for(i=0; i<d; i++){ printf("\n sales %d sales: ",i+1);scanf("%d", &s[i]);}for(i=0; i<d; i++){for(j=i+1; j<d; j++){if(s[i] > s[j]){p= s[i];s[i] = s[j];s[j] = p;}}}printf("\n sales lowest to highest: ");for(i=0; i<d; i++){printf("\nsales %d ", s[i]);}printf("\n highest: %d ",t);getch();} it should be: highest : 2000#include <stdio.h>int main(){int d;int s[20],i, r, p, lg=0,m;char c;printf("Enter number of salesman(max 20): ");scanf("%d", &d);for(i=0; i<d; i++){ printf("\n salesman %d sales: ");scanf("%d", &s[i]);}for(i=0; i<d; i++){for(r=i+1; j<d; r++){if(s[i] > s[r]){p= s[i];s[i] = s[r];s[j] = p;}}}printf("\nsalesman lowest to highest: ");for(i=0; i<d; i++){printf("%d\t", s[i]);}for(i=0;i<m;i++){printf("\n highest sales: %d ",lg);if(lg<=s[i])lg=s[i];break;}getch();} >>> the upper part output should be like this enter number of salesman (max 20): 5 salesman 1 500 salesman 2 300 salesman 3 1000 salesman 4 200 salesman 5 1000 in the lower part the lowest to highest the output should become like this salesman 4 200 salesman 2 300 salesman 1 500 salesman 3 1000 salesman 5 1000 highest total sales : 2000#include<stdio.h> #include<stdlib.h> main() { int m, sum = 0, counter = 0; int first = 2147483647, second = 2147483647, third = 2147483647, min = 2147483647; double average;printf("Enter an int or -1 to stop:\n"); while (1) {scanf_s("\n%d", &m);if (m == -1) { break; } sum = sum + m; counter++; if (m < first first == second first == third ) { third = second; second = first; first = m; } else if (m < second && m !=first ) { third = second; second = m; } else if (m < third && m != second) { third = m; } else if (first == second == third) { printf("min is: %d", first); }}printf("Sum of value: %d \n", sum); average =(double) sum /(double)counter; printf("Avarege is: %.2lf \n", average); printf("First min is: %d\n", first); printf("Second min is: %d\n", second); printf("Third min is: %d\n", third);system("pause"); } can you run this code please2- The factorial n! of a positive integer n is defined as n! = 1*2*3 . .. * (n-1) * n Where 0! = 1 Write a function to calculate the factorial of a number. Argument: A number n of type unsigned int. Returns: The factorial n! of type long double. Write two versions of the function, where the factorial is • calculated using a loop calculated recursively Test both functions by outputting the factorials of the numbers 0 to 20.(Numerical) Write a program that tests the effectiveness of the rand() library function. Start by initializing 10 counters to 0, and then generate a large number of pseudorandom integers between 0 and 9. Each time a 0 occurs, increment the variable you have designated as the zero counter; when a 1 occurs, increment the counter variable that’s keeping count of the 1s that occur; and so on. Finally, display the number of 0s, 1s, 2s, and so on that occurred and the percentage of the time they occurred.#include <stdio.h>void cubeByReference( int *nPtr ); // function prototypeint main( void ){ int number = 5; // initialize number printf("The original value of number is %d", number ); // pass address of number to cubeByReference cubeByReference( &number ); printf("\nThe new value of number is %d\n", number );} // end main void cubeByReference( int *nPtr ){ *nPtr = *nPtr* *nPtr* *nPtr;} passing argument by reference - We modify the code above 1- define a second argument (example "int number2 = 9") and a pointer to it 2- define a second function (addByReference) that adds number2 to number - passing both arguments by reference 3- print-out the result (that is in number) Upload the output and .c code#include <stdio.h> #include <stdlib.h> double f(double x){ return x*x; } double reimannSums(double f(double), double a, double b, int n, double deltax) double sum=0.0, x; int i; printf("n = %d\n",nvoidvoid); for (i=0;i<n;i++){ x= ( (a+i*deltax)+a+(i+1)*deltax))/2; printf("x = %lf\n,x"); sum=sum + f(x); } return sum = (sum * deltax); double integ(double a, double b, double deltax) { double result; int n; n = (b - a) / deltax; return reimannSums(f, a, b, n, deltax); } int main(int argc, char * argv[]) { int a = atoi(argv[1]); int b = atoi(argv[2]); double deltax = atof(argv[3]); printf("result= %lf",integ(a, b, deltax)); } CAN YOU PLEASE CHECK MY WORK I GOT BUILD FAILED ERROR C LANGUAGE I NEED ANSWER IN 30 MIN THANKS<3Given the following code and output:int a[] = {10, 20, 30, 40};cout << a << endl; Example Output3F70#include <stdio.h>int main() { int a = 50; int ans = a++ + ++a + a++ + ++a; printf("%d", ans); return 0;} I neel output of this code.C - Write a program that takes a first name as the input, and outputs a welcome message to that name.13. Given the code char* myFunc (char *ptr, int n) { ptr += n; return ptr; } int main() { char x[100] = Code1; char *y. int n = Code2; y = myFunc (x, n); printf ("%s\n", y); return 0; } where Code1 and Code2 are not given. Which of the following does not give the output ABCDEF Code1 is "XYZABCDEF" and Code2 is 3 Code1 is "ABCABCDEF" and Code2 is 3 Code1 is "ABCDEFXY" and Code2 is 6 Code1 is "ABCDEF" and Code2 is 0SEE MORE QUESTIONS
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