def task(n,m): for i in range(n): s=0 j=i while(j>0): s+=(j%10)**2 j=j//10 if s==m: print(i) task(1000,10)
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- Given code (copy-paste): Problem (see pic): def createList(n): #Base Case/s #ToDo: Add conditions here for base case/s #if <condition> : #return <value> #Recursive Case/s #ToDo: Add conditions here for your recursive case/s #else: #return <operation and recursive call> #remove the line after this once all ToDo is completed return [] def removeMultiples(x, arr): #Base Case/s #TODO: Add conditions here for your base case/s #if <condition> : #return <value> #Recursive Case/s #TODO: Add conditions here for your recursive case/s #else: #return <operation and recursive call> #remove the line after this once you've completed all ToDo return [] def Sieve_of_Eratosthenes(list): #Base Case/s if len(list) < 1 : return list #Recursive Case/s else: return [list[0]] + Sieve_of_Eratosthenes(removeMultiples(list[0], list[1:])) if __name__ == "__main__": n =…Python Language Q2: Ginormous Campus The campus is pretty big. There are n buildings scattered around it, numbered from 0 to n-1. These buildings are so far away from each other that the only way to get from one to another is to take a campus bus. There are m campus bus routes. The i-th one (0 <= i < m) takes you from building u_i to building v_i (but not the other way around). These buses run very frequently. Professor Zingaro is deciding where to hold his CS lectures. He believes a building x is accessible from a building y if you can get from y to x taking at most two buses. For his students’ convenience, he wants to hold his lectures in the most accessible building. Help him out by telling him how many buildings the most accessible building is accessible from. In addition, list all buildings that are the most accessible. Input The first line of the input contains two space-separated integers n and m, denoting the number of buildings and bus routes, respectively. m lines…Outputs of the following: void main () { int M[3]; int *ptr; ptr=M; *ptr=D100; ptr++; *ptr=200; ptr=&M[2]; *ptr=300; for (int n=O; n<3; n++) cout <Modify task2.c to print out the first 50 numbers from the Fibonacci Sequence. task2.c Fibonacci Sequence printing out the first 50 fibonaccinumbers */ #include <stdio.h> int main() { return 0; }Create a task that takes a different number of arguments, each argument representing the number of items in the group. The function should return the number of permissions (combinations) of options you would have if you selected one item per group. Examples combination (2, 3) ➡ 6 -> combination (3, 7, 4) combination (2, 3, 4, 5) - 120 Solve in JS -84def move (A): if len (A) A[0]: more.append(A[i]) less.append(A[i]) return less, more # An example # move([25, 10, 27, 40, 15, 7, 36]) ([10, 15, 7], [27, 40, 36]) Problem 7 Find the running time complexity in terms of of the function "move" in the previous problem.inhinclude Rsing namespace std; int maxResult( ) int maxVal for (int i { for (int j = 0; j <= n 0; i <= n; i += a) i; j += b) %3D %3D float z = (float)(n (i + j)) / (float)(c); if (floor(z) { int x = int y ceil(z)) i / a; j/ b; maxVal = max(maxVal, x + y (int)z); return maxVal; } int main() { max cout << maxResult( ); return 0; 1 } Input Compilation failed due to fellowing ons main.cpp:7:23: error: 'n' was not declared in this scope 7| for (int i = 0; i <= n; i a) main.cpp:7:31: error: 'a was not declared in this scope for (int i = 0; i <= n; i t= a) %3D main.cpp:9:36: error: b' was not declared in this scope | 6 for (int j = 0; j <= n - i; j - b) %3D main.cpp:11:45: error: 'c was not declared in this scopeHi! I get this error message with this code. Can you help me? # Write your solution here def who_won(game_board): # Initialize counters for each player's encircled area player1_area = 0 player2_area = 0 # Iterate through each square in the game board for i in range(len(game_board)): for j in range(len(game_board[i])): # Check if the square is encircled by player 1's game pieces if game_board[i][j] == 1: # Check if all four surrounding squares are also player 1's game pieces if (i > 0 and game_board[i-1][j] == 1) and (i < len(game_board)-1 and game_board[i+1][j] == 1) and (j > 0 and game_board[i][j-1] == 1) and (j < len(game_board[i])-1 and game_board[i][j+1] == 1): player1_area += 1 # Check if the square is encircled by player 2's game pieces elif game_board[i][j] == 2: # Check if all four surrounding squares are also player…Task3: Some of the scientific calculators are capable of identifying errors based on the imbalance of parentheses used in mathematical equations. (4+6)*[7/(9-3)] ==> ()[()] ==> BALANCED (4+6)*7/(9-3)] ==> ()()] ==> IM-BALANCED From the above-mentioned example, we can clearly see that the checking of balance in an equation is simple, which is comparing opening parentheses with the closing ones. As a developer, you are required to write a code that checks an equation for being balanced or not.(HINT: use stack) in c++Add comments in code #include <stdio.h>#include <stdlib.h>/* ADJACENCY MATRIX */int source,V,E,time,visited[20],G[20][20];void DFS(int i){ int j; visited[i]=1; printf(" %d->",i+1); for(j=0;j<V;j++) { if(G[i][j]==1&&visited[j]==0) DFS(j); }}int main(){ int i,j,v1,v2; printf("\t\t\tGraphs\n"); printf("Enter the no of edges:"); scanf("%d",&E); printf("Enter the no of vertices:"); scanf("%d",&V); for(i=0;i<V;i++) { for(j=0;j<V;j++) G[i][j]=0; } /* creating edges :P */ for(i=0;i<E;i++) { printf("Enter the edges (format: V1 V2) : "); scanf("%d%d",&v1,&v2); G[v1-1][v2-1]=1; } for(i=0;i<V;i++) { for(j=0;j<V;j++) printf(" %d ",G[i][j]); printf("\n"); } printf("Enter the source: "); scanf("%d",&source); DFS(source-1); return 0;}1. task: Introduction and basics. And again, Peter Puzzle has found a piece of paper with the following code: int FooBar (A, i) { n= len (A); if (i == n-1) return A; for (j= (i + 1); jUse the template below: def createList(n): #Base Case/s #ToDo: Add conditions here for base case/s #if <condition> : #return <value> #Recursive Case/s #ToDo: Add conditions here for your recursive case/s #else: #return <operation and recursive call> #remove the line after this once all ToDo is completed return [] def removeMultiples(x, arr): #Base Case/s #TODO: Add conditions here for your base case/s #if <condition> : #return <value> #Recursive Case/s #TODO: Add conditions here for your recursive case/s #else: #return <operation and recursive call> #remove the line after this once you've completed all ToDo return [] def Sieve_of_Eratosthenes(list): #Base Case/s if len(list) < 1 : return list #Recursive Case/s else: return [list[0]] + Sieve_of_Eratosthenes(removeMultiples(list[0], list[1:])) if __name__ == "__main__": n = int(input("Enter n: "))…SEE MORE QUESTIONSRecommended textbooks for youDatabase System ConceptsComputer ScienceISBN:9780078022159Author:Abraham Silberschatz Professor, Henry F. 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