Each time we add another bit, what happens to the amount of numbers we can make?

Compare Fibonacci (recursion vs. bottom up)

ipn [44]

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;

}

7 0

2 years ago

SOMEONE PLEASE HELP ME PLEASE HELP ME WITH THIS!!!!!

zavuch27 [327]

Answer:

The last one

Explanation:

I hope this is correct and have a great day

7 0

2 years ago

Read 2 more answers

In a certain computer program, two positive integers are added together, resulting in an overflow error. Which of the following

Airida [17]

The option that best explains why the error occurs is that The program can only use a fixed number of bits to represent integers; the computed sum is greater than the maximum representable value.

<h3>Can programs represent integers?</h3>

An integer value is known to be often listed out in the source code of a program in a way called a sequence of digits that is said to be optionally prefixed with + or −. Note that some programming languages do use other notations, like hexadecimal.

Computers are known to use a a fixed number of bits to show an integer. The most -used bit-lengths for integers are known to be 8-bit, 16-bit, 32-bit or 64-bit.

Learn more about errors from

brainly.com/question/11472659

8 0

2 years ago

Assume that a large number of consecutive IP addresses are available starting at 198.16.0.0 and suppose that two organizations,

Fiesta28 [93]

Answer & Explanation:

An IP version 4 address is of the form w.x.y.z/s

where s = subnet mask

w = first 8 bit field, x = 2nd 8 bit field, y = 3rd 8 bit field, and z = 4th 8 bit field

each field has 256 decimal equivalent. that is

binary denary or decimal

11111111 = 2⁸ = 256

w.x.y.z represents

in binary

11111111.11111111.11111111.11111111

in denary

255.255.255.255

note that 255 = 2⁸ - 1 = no of valid hosts/addresses

there are classes of addresses, that is

class A = w.0.0.0 example 10.0.0.0

class B = w.x.0.0 example 172.16.0.0

class C = w.x.y.0 example 198.16.8.1

where w, x, y, z could take numbers from 1 to 255

Now in the question

we were given the ip address : 198.16.0.0 (class B)

address of quantity 4000, 2000, 8000 is possible with a subnet mask of type

255.255.0.0 (denary) or

11111111.11111111.00000000.00000000(binary) where /s = /16 That is no of 1s

In a VLSM (Variable Length Subnet Mask)

Step 1

we convert the number of host/addresses for company A to binary

4000 = 111110100000 = 12 bit

step 2 (subnet mask)

vary the fixed subnet mask to reserve zeros (0s) for the 12 bit above

fixed subnet mask: 11111111.11111111.00000000.00000000 /16

variable subnet mask: 11111111.11111111.11110000.000000 /20

now we have added 4 1s in the 3rd field to reserve 12 0s

subnet mask: 255.255.16.0 (where the 1s in each field represent a denary number as follows)

1st 1 = 128, 2nd 1 = 64 as follows

1 1 1  1  1 1 1 1

128 64 32 16 8 4 2 1

step 3

in the ip network address: 198.16.0.0/19 (subnet representation) we increment this using 16

that is 16 is added to the 3rd field as follows

That means the ist Valid Ip address starts from

Ist valid Ip add: 198.16.0.1 - 198.16.15.255(last valid IP address)

Company B starts+16: 198.16.16.0 - 198.16.31.255

+16: 198.16.32.0- 198.16.47.255 etc

we repeat the steps for other companies as follows

Company B

Step 1

we convert the number of host/addresses for company B to binary

2000 = 11111010000 = 11 bit

Step 2

vary the fixed subnet mask to reserve zeros (0s) for the 11 bit above

fixed subnet mask: 11111111.11111111.00000000.00000000 /16

variable subnet mask: 11111111.11111111.11111000.000000 /21

now we have added 5 1s in the third field to reserve 11 0s

subnet mask: 255.255.8.0 (where the 1s in each field represent a denary number as follows)

1st 1 = 128, 2nd 1 = 64 as follows

1 1 1 1 1  1 1 1

128 64 32 16 8  4 2 1

Step 3

Starting from after the last valid Ip address for company A

in the ip network address: 198.16.16.0/21 (subnet representation) we increment this using 8

That means the ist Valid Ip address starts from

Ist valid Ip add: 198.16.16.1 - 198.16.23.255(last valid IP address)

Company C starts +16: 198.16.24.0- 198.16.31.255

+16: 198.16.32.0- 198.16.112.255 etc

Company C

Step 1

we convert the number of host/addresses for company C to binary

4000 = 111110100000 = 12 bit

Step 2

vary the fixed subnet mask to reserve zeros (0s) for the 12 bit above

fixed subnet mask: 11111111.11111111.00000000.00000000 /16

variable subnet mask: 11111111.11111111.11110000.000000 /20

now we have added 4 1s in the 3rd field to reserve 12 0s

subnet mask: 255.255.16.0 (where the 1s in each field represent a denary number as follows)

1st 1 = 128, 2nd 1 = 64 as follows

1 1 1 1 1 1 1 1

128 64 32 16 8 4 2 1

Step 3

Starting from after the last valid ip address for company B

in the ip network address: 198.16.24.0/20 (subnet representation) we increment this using 16

That means the ist Valid Ip address starts from

Ist valid Ip add: 198.16.24.1 - 198.16.39.255(last valid IP address)

Company C starts +16: 198.16.40.0- 198.16.55.255

+16: 198.16.56.0- 198.16.71.255 etc

Company D

Step 1

we convert the number of host/addresses for company D to binary

8000 = 1111101000000 = 13 bit

Step 2

vary the fixed subnet mask to reserve zeros (0s) for the 13 bit above

fixed subnet mask: 11111111.11111111.00000000.00000000 /16

variable subnet mask: 11111111.11111111.11100000.000000 /19

now we have added 3 1s in the 3rd field to reserve 13 0s

subnet mask: 255.255.32.0 (where the 1s in each field represent a denary number as follows)

1st 1 = 128, 2nd 1 = 64 as follows

1 1  1  1 1 1 1 1

128 64 32  16 8 4 2 1

Step 3

Starting from after the last valid ip address for company C

in the ip network address: 198.16.40.0/20 (subnet representation) we increment this using 32

That means the ist Valid Ip address starts from

Ist valid Ip add: 198.16.40.1 - 198.16.71.255(last valid IP address)

Company C starts +16: 198.16.72.0- 198.16.103.255

+16: 198.16.104.0- 198.16.136.255 etc

5 0

3 years ago

You need to install a 32-bit application on a 32-bit version of windows 10. In which default directory will the application be i

luda_lava [24]

Answer:

%systemdrive%\Program Files

7 0

2 years ago