add comments in code like what for loop doing what and other functions #include #include /* 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
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add comments in code like what for loop doing what and other functions
#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;
}
Step by step
Solved in 3 steps with 1 images
- #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 LEvoid listEmployees (void) { for (int i=0; i 10000. Make a guess about why the comparison function takes 2 struct Employee parameters (as opposed to struct Employee *) **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;}
- briefly comment the 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;}Not allowed to add variables in brackets next to originIndex * @return index of the point that is closest to the origin, which is (0, 0) In * case of a tie, return the lowest index */ public int closestToOriginIndex() {C++ A robot is initially located at position (0; 0) in a grid [?5; 5] [?5; 5]. The robot can move randomly in any of the directions: up, down, left, right. The robot can only move one step at a time. For each move, print the direction of the move and the current position of the robot. If the robot makes a circle, which means it moves back to the original place, print "Back to the origin!" to the console and stop the program. If it reaches the boundary of the grid, print \Hit the boundary!" to the console and stop the program. A successful run of your code may look like:Down (0,-1)Down (0,-2)Up (0,-1)Left (-1,-1)Left (-2,-1)Up (-2,0)Left (-3,0)Left (-4,0)Left (-5,0)Hit the boundary! or Left (-1,0)Down (-1,-1)Right (0,-1)Up (0,0)Back to the origin! About: This program is to give you practice using the control ow, the random number generator, and output formatting. You may use <iomanip> to format your output. You may NOT use #include "stdafx.h".
- #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<A/ find code optimization method to this code and then find type to this code 1- For(i=1;iint a[10] = {0,1,2,3,4,5,6,7,8,9};int *m = &a[0];int *p = &a[5];int *q = &a[1]; p = (int) m + (int) p - (int) q; what is the value of of *pQuestion 37 public static void main(String[] args) { Dog[] dogs = { new Dog(), new Dog()}; for(int i = 0; i >>"+decision()); } class Counter { private static int count; public static void inc() { count++;} public static int getCount() {return count;} } class Dog extends Counter{ public Dog(){} public void wo(){inc();} } class Cat extends Counter{ public Cat(){} public void me(){inc();} } The Correct answer: Nothing is output O 2 woofs and 5 mews O 2 woofs and 3 mews O 5 woofs and 5 mews OQ1 #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…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 stepsSEE MORE QUESTIONSRecommended textbooks for youDatabase System ConceptsComputer ScienceISBN:9780078022159Author:Abraham Silberschatz Professor, Henry F. Korth, S. SudarshanPublisher:McGraw-Hill EducationStarting Out with Python (4th Edition)Computer ScienceISBN:9780134444321Author:Tony GaddisPublisher:PEARSONDigital Fundamentals (11th Edition)Computer ScienceISBN:9780132737968Author:Thomas L. FloydPublisher:PEARSONC How to Program (8th Edition)Computer ScienceISBN:9780133976892Author:Paul J. Deitel, Harvey DeitelPublisher:PEARSONDatabase Systems: Design, Implementation, & Manag…Computer ScienceISBN:9781337627900Author:Carlos Coronel, Steven MorrisPublisher:Cengage LearningProgrammable Logic ControllersComputer ScienceISBN:9780073373843Author:Frank D. PetruzellaPublisher:McGraw-Hill EducationDatabase System ConceptsComputer ScienceISBN:9780078022159Author:Abraham Silberschatz Professor, Henry F. Korth, S. SudarshanPublisher:McGraw-Hill EducationStarting Out with Python (4th Edition)Computer ScienceISBN:9780134444321Author:Tony GaddisPublisher:PEARSONDigital Fundamentals (11th Edition)Computer ScienceISBN:9780132737968Author:Thomas L. FloydPublisher:PEARSONC How to Program (8th Edition)Computer ScienceISBN:9780133976892Author:Paul J. Deitel, Harvey DeitelPublisher:PEARSONDatabase Systems: Design, Implementation, & Manag…Computer ScienceISBN:9781337627900Author:Carlos Coronel, Steven MorrisPublisher:Cengage LearningProgrammable Logic ControllersComputer ScienceISBN:9780073373843Author:Frank D. PetruzellaPublisher:McGraw-Hill Education