int sum, k, i, j; int x[4][4]={1,2,3,4},{5,6,7,8},{9,8,7,3},{2,1,7,1}; sum=x[0][0]; for (k=1; k<=3;k++) sum+=x[k] [k]; Give the value in sum after the statements are executed:
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![int sum, k, i, j;
int x[4] [4]={1,2,3,4}, {5,6,7,8},{9,8,7,3},{2, 1,7,1};
sum=x[0] [0];
for (k=1; k<=3;k++)
sum+=x[k] [k];
Give the value in sum after the statements are executed:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F521df3ae-8e5f-4edb-bcc7-9453c7ca96c5%2Ff6094e0d-6a3f-4f15-b6da-130d0145fdde%2Fvngu9q_processed.jpeg&w=3840&q=75)
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- #include for(int i = 0; i { int main() for(int j = 0; j< 2; j++) { { int matrix[2][2] = { printf(" %d", matrix[i][j]); } {2,3,}, //rowo printf("\n"); } {5,7}//row1 getch(); }; } printf("\n Resultant \n"); Try to execute, screenshot the resul and try to explain the sequence of the program. Post it in space-ners provided. FRAMEWORK HTPE ADACHE e OrcE outATION roR TECHO A LEint X[10]={2,0,6,11,4,5,9,11,-2,-1); From the code above, what is the value of X[8] ?int c[2][3]; int a[2][3] ={ {5}, {1,2} }; int b[2][3] = { 2,4,6,8} for( int i=0; i<2 ;i++)5. for( int j=0; j<3 ; j++) c[i] [j] = 3 * ( a[i] [j] + b[i] [j] ); program with c++
- JAVA Programming Write a function that returns true if you can partition an array into one element and the rest, such that this element is equal to the product of all other elements excluding itself. Examples canPartition ([2, 8, 4, 1]) → true // 8 = 2 x 4 x 1 canPartition ([-1, -10, 1, -2, 20]) → false canPartition ([-1, -20, 5, -1, -2, 2]) trueint main() //default function for call { int a[100],n,i,j; for (int i = 0; i < n; i++) //Loop for ascending ordering { for (int j = 0; j < n; j++) //Loop for comparing other values { if (a[j] > a[i]) //Comparing other array elements { int tmp = a[i]; //Using temporary variable for storing last value a[i] = a[j]; //replacing value a[j] = tmp; //storing last value } } } printf("\n\nAscending : "); //Printing message for (int i = 0; i < n; i++) //Loop for printing array data after sorting { printf(" %d ", a[i]); } } Need to transform this C code to MIPS Assembly Language Code simple stepsint main() { int inStock[10][4]; int alpha[20]; int beta[20]; int gamma[4] = {11, 13, 15, 17}; int delta[10] = {3, 5, 2, 6, 10, 9, 7, 11,1, 8};. . .} a. Write the definition of the function setZero that initializes any onedimensional array of type int to 0. b. Write the definition of the function inputArray that prompts the user to input 20 numbers and stores the numbers into alpha. c. Write the definition of the function doubleArray that initializes the elements of beta to two times the corresponding elements in alpha. Make sure that you prevent the function from modifying the elements of alpha. d. Write the definition of the function copyGamma that sets the elements of the first row of inStock to gamma and the remaining rows of inStock to three times the previous row of inStock. Make sure that you prevent the function from modifying the elements of gamma. e. Write the definition of the function copyAlphaBeta that stores alpha into the first five rows of inStock and beta into the…
- typedef struct { short data[4];} MatrixElement; void copy_matrix(MatrixElement m1[], MatrixElement m2[], int ROWS, int COLS) { int i, j, k; for (i = 0; i < ROWS; i++) { for (j = 0; j < COLS; j++) { for (k = 0; k < 4; k++) { m1[i*COLS+j].data[k] = m2[i*COLS+j].data[k]; } } }} void copy_matrix_transpose(MatrixElement m1[], MatrixElement m2[], int ROWS, int COLS) { int i, j, k; for (i = 0; i < ROWS; i++) { for (j = 0; j < COLS; j++) { for (k = 0; k < 4; k++) { m1[i*COLS+j].data[k] = m2[j*ROWS+i].data[k]; } } }} You can assume the following conditions: The matrix m1 is allocated at memory address 0, and matrix m2 immediately follows it. Indices i, j, and k are kept in registers. ROWS and COLS are constants. The cache is initially empty before the function call. The cache is write-back (i.e., only writes back to memory when a line is…Debug Program3: int element(int arr[], int num) { in i, ele; ele = arr[0]; for (i = 1; i < num; i++) if (arr[i] > ele) ele = arr[i]; return ele; } main() { arr[] = {1, 24, 145, 20, 8, -101, 300}; int n = sizeof(arr)/sizeof(arr[0]); printf("element of array is %d", element(arr[], n)); } 24 145 -101 300 Program4: main() { int m,n; for(i 0; i << 5; i++) { printf("\t\t\t\t"); for(j 0; j << 5; j+++) printf("*"); printf("/n"); } Program 5: int func(int) main() { int num = 7, c; int f= func(num); printf("\n\n%d of %d\n\n", num, f); } int func(aj); { if(aj=1 aj=0) return 1; else return (aj*fun(aj-1); } 49 5040 7117 none of…Java Objective:Design and implement simple matrix manipulation techniques.Details:Your java program should use 2D arrays to implement simple matrix operations.Your program should do the following:• Read the number of rows and columns of a matrix M1 from the user. Use an input validation loop to make sure the values are greater than 0. • Read the elements of M1 in row major order• Print M1 to the console; make sure you format as a matirx• Repeat the previous steps for a second matrix M2• Create a matrix M3 that is the transpose of M1 and print it to the console• Check if M1 and M2 can be added (should have the same dimensions). If possible, add M1 and M2 and print the result to the console. Otherwise print an error message.• Multiply M1 and M2 if possible and print to the console. If the matrices cannot be multiplied, print an error message. Implementation requirements:• Use a helper method for reading a positive integer using an input validation loop.• Use a helper method for printing…
- A = {1,2,3} B = {1,2,3} C = {5,6,7} D = {0,1,2,3,4,5} Is A = C?#include <stdio.h> int arrC[10] = {0}; int bSearch(int arr[], int l, int h, int key); int *joinArray(int arrA[], int arrB[]) { int j = 0; if ((arrB[0] + arrB[4]) % 5 == 0) { arrB[0] = 0; arrB[4] = 0; } for (int i = 0; i < 5; i++) { arrC[j++] = arrA[i]; if (arrB[i] == 0 || (bSearch(arrA, 0, 5, arrB[i]) != -1)) { continue; } else arrC[j++] = arrB[i]; } for (int i = 0; i < j; i++) { int temp; for (int k = i + 1; k < j; k++) { if (arrC[i] > arrC[k]) { temp = arrC[i]; arrC[i] = arrC[k]; arrC[k] = temp; } } } for (int i = 0; i < j; i++) { printf("%d ", arrC[i]); } return arrC; } int bSearch(int arr[], int l, int h, int key) { if (h >= l) { int mid = l + (h - l) / 2; if…8Consider the following ANSI C function: int SimpleFunction (int Y[ ], int n, int rx) { int total = Y[0], looplndex; for (looplndex = 1; looplndex <= n - 1; looplndex++) total = x x total + Y[looplndex]; return total; Let Z be an array of 10 elements with Z[i] = 1, for all i such that 0 sis 9. The value returned by SimpleFunction (Z, 10, 2) is (100 01