It will probably cost extra
Answer:
Check the explanation
Explanation:
#include <iostream>
using namespace std;
void hex2dec(string hex_num){
int n = 0;
//Loop through all characters in string
for(int i=0;i<hex_num.size();i++){
//take ith character
char c = hex_num[i];
//Check if c is digit
if(c>='0' && c<='9'){
n = 16*n + (c-48);
}
//Convert c to decimal
else{
n = 16*n + (c-55);
}
}
cout<<hex_num<<" : "<<n<<endl;
}
int main()
{
hex2dec("EF10");
hex2dec("AA");
return 0;
}
The Output can be seen below :
Answer:
see explaination
Explanation:
#include <iostream>
#include <string>
using namespace std;
class LinkedList{
class Node{
public :
int data;
Node* next;
Node(int data){
this->data = data;
next = NULL;
}
};
public :
Node *head;
LinkedList(){
this->head = NULL;
}
void insert(int d){
Node* new_node = new Node(d);
new_node->next = head;
head = new_node;
}
// sort the list with selection sort algorithm.
// Pick the smallest element in the unsorted array and place in the first element in the unsorted.
void sort_list(){
if (head == NULL){
return;
}
Node* current = head;
while (current->next != NULL){
Node* min_node = current;
Node* traverse = current->next;
while(traverse != NULL){
if(traverse->data < min_node->data){
min_node = traverse;
}
traverse = traverse->next;
}
int temp = current->data;
current->data = min_node->data;
min_node->data = temp;
current = current->next;
}
}
void print_list(){
Node* current = head;
while(current !=NULL){
cout<<current->data<<" ";
current = current->next;
}
cout<<"\n";
}
};
int main(){
LinkedList ll;
for(int i=0;i<10;i++){
ll.insert(i);
}
ll.print_list();
cout<<"*******************************************\n";
ll.sort_list();
ll.print_list();
cout<<"*******************************************\n";
}
Answer:
By backing up the important information from the laptop computers to OneDrive, if a hard drive failure is to occur, assuming the salespeople have internet, they may access OneDrive and download any and all applicable data, resolving the issue.
Answer: True
Explanation:
Subset sum problem and Knapsack problem can be solved using dynamic programming.
In case of Knapsack problem there is a set of weights associative with objects and a set of profits associated with each object and a total capacity of knapsack let say C. With the help of dynamic programming we try to include object's weight such that total profit is maximized without fragmenting any weight of objects and without exceeding the capacity of knapsack, it is also called as 0/1 knapsack problem.
Similar to knapsack problem, in subset sum problem there is set of items and a set of weights associated with the items and a capacity let say C, task is to choose the subset of items such that total sum of weights associated with items of subset is maximized without exceeding the total capacity.
On the basis of above statements we can say that subset sum problem is generalization of knapsack problem.
