Answer:
A.Sam.
Explanation:
We are using Deque interface which is sub type of Queue interface. Deque supports insertion and deletion from both ends front and end.So it can be used as a queue and stack also.
In this question we have inserted Jack at the front first.Then Rudy at the back then larry also at the tail.Now we have added sam at the front then nothing is added to the front.So the answer is Sam.
Answer: MIB
Explanation:
management information base(MIB) is a database which is to be used to allow the network management tool to interpret the new device and control it using SNMP.
MIB is a virtual database of all the network object so that network management tool can keep a track of it.
Answer:
I'm pretty sure it's photogenic drawing.
Answer:
#include <iostream>
#include <ctime>
#include <cstdlib>
using namespace std;
int menu();
int doubleIt(int);
int getPower(int,int);
int reverse(int);
int power(int);
int sum(int);
int twoDigit(int);
int threeDigit(int);
bool prime(int);
int main()
{int n;
srand(time(0));
n=rand()%90+10;
do
{
cout<<"The number is: "<<n<<endl<<endl;
switch(menu())
{case 1: n=doubleIt(n);
break;
case 2: n=reverse(n);
break;
case 3: n=power(n);
break;
case 4: n=sum(n);
break;
case 5: n=twoDigit(n);
break;
case 6: n=threeDigit(n);
break;
case 7: return 0;
}
if(prime(n))
cout<<n<<" is prime\n";
else
cout<<n<<" is not prime\n\n";
if(n<10)
n+=10;
}while(true);
}
int doubleIt(int n)
{return n*2;
}
int power(int n)
{int p;
cout<<"Enter 2, 3 or 4th power: ";
cin>>p;
while(p<2||p>4)
{cout<<"invalid entry\n";
cout<<"Enter 2, 3 or 4th power: ";
cin>>p;
}
return getPower(n,p);
}
int reverse(int n)
{int newnum=0;
while(n>0)
{newnum=newnum*10+n%10;
n=n/10;
}
return newnum;
}
int sum(int n)
{int sum=0;
while(n>0)
{sum=sum+n%10;
n=n/10;
}
return sum;
}
int twoDigit(int n)
{int d1,d2;
if(n<100)
{d1=n%10;
d2=n/10;
return getPower(d1,d2);
}
return n;
}
int threeDigit(int n)
{int d1,n2;
if(n>99)
{d1=n%10;
if(d1<=4)
{n2=n/10;
return getPower(n2,d1);
}
}
return n;
}
bool prime(int n)
{int i;
for(i=2;i<n-1;i++)
if(n%i==0)
return false;
return true;
}
int getPower(int x,int y)
{int ans,i;
if(y==0)
return 1;
else
{ans=x;
for(i=2;i<=y;i++)
ans=ans*x;
return ans;
}
}
int menu()
{int choice=8;
while(choice<1||choice>7)
{cout<<"1. double the number\n";
cout<<"2. reverse the digit of the number\n";
cout<<"3. raise the number to the power of 2,3, or 4\n";
cout<<"4. sum thee digits of the number.\n";
cout<<"5. if the number is a two digit number\n then raise the first digit to the power of the second digit\n";
cout<<"6. If the number is a three digit number\n";
cout<<" and the last digit is less than or equal to 4\n then raise the first two digits to the power of the last digit.\n";
cout<<"7. exit\n";
cin>>choice;
}
return choice;
}
The answer for the 1st blank is text.