Answer:
Explanation:
Since all of the items in the array would be integers sorting them would not be a problem regardless of the difference in integers. O(n) time would be impossible unless the array is already sorted, otherwise, the best runtime we can hope for would be such a method like the one below with a runtime of O(n^2)
static void sortingMethod(int arr[], int n)
{
int x, y, temp;
boolean swapped;
for (x = 0; x < n - 1; x++)
{
swapped = false;
for (y = 0; y < n - x - 1; y++)
{
if (arr[y] > arr[y + 1])
{
temp = arr[y];
arr[y] = arr[y + 1];
arr[y + 1] = temp;
swapped = true;
}
}
if (swapped == false)
break;
}
}
Answer:
by it scan properly like you use Camara
when we use camara and it have record time like that it also have that type of scan data so that the computer use
sorry I don't no other thing and I an sorry for the very short answer
Answer: 32 bit number
Explanation:
The IP address basically contain 32 bit number as due to the growth of the various internet application and depletion of the IPV4 address. The IP address basically provide two main function is that:
- The location addressing
- The network interface identification
The IP address are basically available in the human readable format. The IPV6 is the new version of the IP address and its uses 128 bits.
Answer:
Explanation:
#include <iostream>
using namespace std;
int costdays(int);
int costhrs(int,int);
int main()
{
int dd,hh,mm,tmph,tmpd,tmpm=0;
int pcost,mcost=0;
cout<<"Enter Parking time" << endl;
cout<<"Hours: ";
cin>>hh;
cout<<"Minutes: ";
cin>>mm;
if (mm>60)
{
tmph=mm/60;
hh+=tmph;
mm-=(tmph*60);
}
if (hh>24)
{
tmpd=hh/24;
dd+=tmpd;
hh-=(tmpd*24);
}
if ((hh>4)&&(mm>0))
{
pcost+=costdays(1);
}
else
{
mcost=costhrs(hh,mm);
}
cout<<"Total time: ";
if (dd>0)
{
cout<<dd<<"days ";
pcost+=costdays(dd);
}
pcost+=mcost;
cout<<hh<<"h "<<mm<<"mins"<<endl;
cout<<"Total Cost :"<<pcost<<"Won";
return 0;
}
int costdays(int dd)
{
return(dd*25000);
}
int costhrs(int hh,int mm)
{
int tmpm, tmp=0;
tmp=(hh*6)*1000;
tmp+=(mm/10)*1000;
tmpm=mm-((mm/10)*10);
if (tmpm>0)
{
tmp+=1000;
}
return(tmp);
}
