All categories

3 years ago

If i'm wanting to use hydra on linux to crack a password and the issue regarding hashes occurs, what shall i do?

Computers and Technology

1 answer:

Inessa05 [86]3 years ago

7 0

Give up do research and then try again

You might be interested in

Write a C++ program that reads from the standard input and counts the number of times each word is seen. A word is a number of n

Rufina [12.5K]

Answer:

#include <stdio.h>

#include <string.h>

#include <ctype.h>

struct detail

{

char word[20];

int freq;

};

int update(struct detail [], const char [], int);

int main()

{

struct detail s[10];

char string[100], unit[20], c;

int i = 0, freq = 0, j = 0, count = 0, num = 0;

for (i = 0; i < 10; i++)

{

s[i].freq = 0;

}

printf("Enter string: ");

i = 0;

{

fflush(stdin);

c = getchar();

string[i++] = c;

} while (c != '\n');

string[i - 1] = '\0';

printf("The string entered is: %s\n", string);

for (i = 0; i < strlen(string); i++)

{

while (i < strlen(string) && string[i] != ' ' && isalnum(string[i]))

{

unit[j++] = string[i++];

}

if (j != 0)

{

unit[j] = '\0';

count = update(s, unit, count);

j = 0;

}

printf("*****************\nWord\tFrequency\n*****************\n");

for (i = 0; i < count; i++)

{

printf("%s\t %d\n", s[i].word, s[i].freq);

if (s[i].freq > 1)

{

num++;

}

printf("The number of repeated words are %d.\n", num);

return 0;

}

int update(struct detail s[], const char unit[], int count)

{

int i;

for (i = 0; i < count; i++)

{

if (strcmp(s[i].word, unit) == 0)

{

s[i].freq++;

return count;

}

/*If control reaches here, it means no match found in struct*/

strcpy(s[count].word, unit);

s[count].freq++;

/*count represents the number of fields updated in array s*/

return (count + 1);

}

4 0

3 years ago

Host A sends real time voice to Host B over a packet switched network. Host A first converts the analogue voice signal to a digi

DiKsa [7]

Answer:

Time required is 0.007 s

Explanation:

As per the question:

Analog signal to digital bit stream conversion by Host A =64 kbps

Byte packets obtained by Host A = 56 bytes

Rate of transmission = 2 Mbps

Propagation delay = 10 ms = 0.01 s

Now,

Considering the packets' first bit, as its transmission is only after the generation of all the bits in the packet.

Time taken to generate and convert all the bits into digital signal is given by;

t = $\frac{Total\ No.\ of\ packets}{A/D\ bit\ stream\ conversion}$

t = $\frac{56\times 8}{64\times 10^{3}}$ (Since, 1 byte = 8 bits)

t = 7 ms = 0.007 s

Time Required for transmission of the packet, t':

$t' = \frac{Total\ No.\ of\ packets}{Transmission\ rate}$

$t' = \frac{56\times 8}{2\times 10^{6}} = 2.24\times 10^{- 4} s$

5 0

2 years ago

Which section of a textbook contains charts, lists, documents, or other material related to the subject of the book

Anton [14]

The apendix is the place to look

3 0

2 years ago

If someone you knew was considering installing Smarthome technology, would you advise that person to install it or not? How woul

jeka94

Is this helpful? I would research extensively

5 0

3 years ago

What are dividends? A

Scilla [17]

<span>a number to be divided by another number.</span>

8 0

2 years ago

Read 2 more answers