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);
}
Answer:
G=(A^3 + B) * 4-c^2
To calculate G if A=3, b=4 and C=5
cobegin
p1: A^3 = 27
P2:C^2 =25
P3: p1 +B =31
P4:P3*4 =124
P5:P4-P2 =99
coend
Explanation:
One operation is performed at a time, and on the basis of priority as we have only one processor.
Answer: customers who placed orders, orders that are attached to customers or C E
Explanation: on edg
