1. int Foo (int a[]) { 2. let n = A.size(); if (n=1) return A[1]; 3. for (i = 1 to n) { 44. print hi 6. } //end for 7. int x=foo (A[1..n/y]); B. 9. 12.) int 2-foo (A[1..2n/y]); return x+s; Analyze line by line, find the recurrence and use substitution to try to prove that the code is O(n)
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- 1 int[] a = new int[n]; 2 a[0] = 0; a[1] = 1; 3 int x = 0; 4 for (int i = 2; i < n; i++) { 5 x = 3 * (i-1) * (i-1); 6 x += 3 * i - 2; 7 a[i] = a[i-1] + x; 8 } The code can be simplified by replacing lines 5, 6, 7 with a single line of code. Write the single line of code below:b) int [] num = new int [6]; for (int i = 1; i 2) { num [i] = 2 * num[i]-num[i-1]; System.out.println( "num [" + i + ") =" + num [i]);extern "C" int f(int *,int,int); int a[2][2] = {{11,12},{21,22}}; void setup(){ Serial.begin(115200); while(!Serial); delay(500); int *arr = (int*)a; Serial.println(f(arr,2,2)); } void loop(){ } .global f f: ldr r3,[r0] // get first element mov r2,#0add r2,r3 mov r1, #3 lp: add r0,#4 // add next elementldr r3,[r0]add r2,r3sub r1,#1 bgt lp mov r0,r2 bx lr Describe what operation is being performed in this code
- #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] ?using namespace std; int main() int i, m=0,n=4; float arr[100] (55,66,88,1); for(i=0; i#include using namespace std; int main() (double degree [6] [2] = {30,40,10,70, 20, 30, 60, 70, 30, 10, 10, 85); int buffer=degree[0][0]; for (int i=1;ibuffer) buffer=degree[i][0]; cout using namespace std; int main() { int j=1; while (j<=10) if ((j%3)==0) cout<int 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 steps// Assume all libraries are included void QQ(int n); // int main () { // Random questions QQ (5) ; 3 4 6. 7 8. 9. 10 return 0; 11 } // void QQ(int n) 12 13 14 { if(n >= 1) { 15 16 17 cout << n; QQ (--n); } else 18 19 20 21 cout << n; 22 } // 23Q1 #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…Consider the following code segment that includes a method called multiply. The multiply method is a sample method for the multiplication of an n by m array and an m by p array. Suppose the multiply method is invoked as shown after replacing the element indicated as x by a number. What number should replace x so that the b3[0][0] = 9 after the execution of b3 = multiply(b1,b2)? Note that the initializeArray method puts a O to every cell of the array passed to initializeArray. *#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…SEE MORE QUESTIONSRecommended textbooks for youDatabase System ConceptsComputer ScienceISBN:9780078022159Author:Abraham Silberschatz Professor, Henry F. Korth, S. 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