t is the output of following section of a program? [10] = {1,2,3,x,5,6,7,8,9.y); = 0, s = 0; e(i<10) s=s+a[i]; i=i+3; <<<<
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![What is the output of following section of a program?
int a[10] = {1,2,3,x,5,6,7,8,9,y);
int i = 0, s = 0;
while(i<10)
{
}
cout<<s<<endl;
s = s + a[i];
i=i+3;
x+8+y
x+7+y
O x
OY](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0bb569e8-9bf0-4abd-bd47-801dd9a4b33e%2Febc83bab-9b71-4bd6-8f81-1bd511714216%2Fh5p9r3m_processed.png&w=3840&q=75)
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- #include using namespace std; const int y = 1; int main () { int static y = 2; int i = 3, j = 4, m = 5, n = 6; %3D int a = [] (int x, int i = 1) { return x * i; } (y, 3) ; %3D int b = [=] (int x) { return [=] (int b) { return b + j; } (x) % 7; } (a); int c = [=] (int x) mutable ->int { m = 6; return [6] (int j) mutable { y = a * b; return y / j; } (x) -m; } (b) ; cout << a << endl; cout << b << endl; cout << c << endl; cout << m << endl; cout << y < endl; return 0; This program outputs 5 lines. What are they? Please explain your work and your answer.CFG: Example 1 • Draw the CFG for the following code: int f(int n){ } int m = n* n; if (n < 0) else return 0; return m;#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> struct Single { int num; }; void printSingle(int f) { int binaryNum[33]; int i = 0; while(f>0) { binaryNum[i] = f % 2; f = f/2; i++; } for (int j=i-1; j>= 0; j--) { printf("%d",binaryNum[j]); } } int main() { struct Single single; single.num = 33; printf("Number: %d\n",single.num); printSingle(single.num); return 0; }Void Do1 (int: &, a. int &b) { a = 5; a = a + b; b = a + 2; } Int main() { Int x = 10; Do1 (x,x); Cout << x << endl; } The output of this program is(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.
- C programming The field of astronomy has been significantly advanced through the use of computer technology. Algorithms can automatically survey digital images of the night sky, looking for new patterns. For this problem, you should write such an analysis program which counts the number of stars visible in an bitmap image. An image consists of pixels, and each pixel is either black or white (represented by the characters # and -, respectively). All black pixels are considered to be part of the sky, and each white pixel is considered to be part of a star. White pixels that are adjacent vertically or horizontally are part of the same star. Input Each test case begins with a line containing a pair of integers 1≤m, n≤100. This is followed by mm lines, each of which contains exactly nn pixels. Input contains at least one and at most 50 test cases, and input ends at the end of file. Output For each case, display the case number followed by the number of stars that are visible in the…int sum = 0; for (int i 0; i < 5; i++){ sum += i; } cout << sum;Consider the following pseudo code, Method func() { PRINT “This is recursive function" func() } Method main( { func() } What will happen when the above snippet is executed?#include <stdio.h> int main(){ int arr[10]; int i; for (i=0; i<10; i++){ arr[i] = i; } for (i=0; i<10; i++){ printf("arr[%d] = %d\n", i, arr[i]); } return 0;} Copy the code and modify it as follows: Add another separate array of 10 integers. The following code parts are to be added (in the same order as specified) after the printing of the values of the first array:(1) Copy the content of the first array to the second array. Add 10 to each of the values of the second array. Make use of a single loop to achieve this part.(2) Print each element of the second array. Make use of another loop for this part#include #include #include int main() { int i, j; printf("Columns | JIn"); for (i = 1; i< 4; ++i) { printf("Outer %6d\n", i); for (j = 0; j< i; ++j) { printf(" Inner%10d\n", j); } } /* heading of outer for loop */ /* heading of inner loop */ %3D return (0); } Create a new code modifying it to use "while loop" instead of the "for loop"SEE MORE QUESTIONS