"Hackers" who work for companies to expose security flaws are definitely good, they're helping protect systems, often referred to as "white-hat" hackers. Many companies hire people like these or offer bounties for such people who can find vulnerabilities. One could say that even without permission, people who hack with the intent of demonstrating holes in security or improving a service could be considered good.
However, I don't believe that as firstly, they are breaking the law, and secondly, they are stealing data and invading people's privacy. One might as well break into somebody's house to say their security is bad!
Answer:
C++ code explained below
Explanation:
#include<bits/stdc++.h>
#include <iostream>
using namespace std;
int FiboNR(int n)
{
int max=n+1;
int F[max];
F[0]=0;F[1]=1;
for(int i=2;i<=n;i++)
{
F[i]=F[i-1]+F[i-2];
}
return (F[n]);
}
int FiboR(int n)
{
if(n==0||n==1)
return n;
else
return (FiboR(n-1)+FiboR(n-2));
}
int main()
{
long long int i,f;
double t1,t2;
int n[]={1,5,10,15,20,25,30,35,40,45,50,55,60,65,70,75};
cout<<"Fibonacci time analysis ( recursive vs. non-recursive "<<endl;
cout<<"Integer FiboR(seconds) FiboNR(seconds) Fibo-value"<<endl;
for(i=0;i<16;i++)
{
clock_t begin = clock();
f=FiboR(n[i]);
clock_t end = clock();
t1=double(end-begin); // elapsed time in milli secons
begin = clock();
f=FiboNR(n[i]);
end = clock();
t2=double(end-begin);
cout<<n[i]<<" "<<t1*1.0/CLOCKS_PER_SEC <<" "<<t2*1.0/CLOCKS_PER_SEC <<" "<<f<<endl; //elapsed time in seconds
}
return 0;
}
<span>If a DNS packet using UDP is lost this problem must be handled at the application layer. If an answer is not received after some period of time the specific application may choose to resend the DNS query. I can go much more into detail but this should answer your question. If you need more detail just ask.</span>
Answer:
Following is the program in C language :
#include <stdio.h> // header file
#define n 5 // macro
int main() main function
{
int a[n],k1; // variable and array declaration
printf("Enter the element:\n");
for(k1=0;k1<n;++k1) //iterating the loop
{
scanf("%d",&a[k1]);//Read the values by user
}
printf("Output in Reverse Order:\n");
for(k1=n-1;k1>=0;--k1)//iterating the loop
{
printf(" %d ",a[k1]); //Display the values
}
return 0;
}
Output:
Enter the element:
4
3
45
67
89
Output in Reverse Order: 89 67 45 3 4
Explanation:
Following is the description of the program
- Define a macro "n" with value 5 after the header file.
- Declared an array "a" and defined the size of that array by macro i.e "n".
- Read the value by the user by using scanf statement in the array "a"
- Finally In the last for loop display the values of array "a" by space.
