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Gwar [14]
3 years ago
12

Targeted attacks are easier and take less time and effort than attacks on targets of opportunity.

Computers and Technology
1 answer:
Mnenie [13.5K]3 years ago
8 0

Answer:

False

Explanation:

Targeted attacks are usually harder because if someone has a password that is 1234 and you try it on 100 computers, you most likely will get someones password. If you are targetting a select computer with a strong password it is much harder to brute force or guess.

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At what point in a vulnerability assessment would an attack tree be utilized?
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3 years ago
Given two character strings s1 and s2. Write a Pthread program to find out the number of substrings, in string s1, that is exact
anyanavicka [17]

Answer:

Explanation:

#include <stdlib.h>

#include <stdio.h>

#include <string.h>

#define MAX 1024

int total = 0 ;

int n1, n2;

char s1, s2;

FILE fp;

int readf(FILE fp)

{

 if((fp=fopen("strings.txt", "r"))==NULL) {

   printf("ERROR: can’t open string.txt!\n");

   return 0;

 }

 s1=(char)malloc(sizeof(char)MAX);

 if(s1==NULL) {

   printf("ERROR: Out of memory!\n");

   return 1;

 }

 s2=(char)malloc(sizeof(char)MAX);

 if(s1==NULL) {

   printf("ERROR: Out of memory!\n");

   return 1;

 }

 /* read s1 s2 from the file */

 s1=fgets(s1, MAX, fp);

 s2=fgets(s2, MAX, fp);

 n1=strlen(s1); /* length of s1 */

 n2=strlen(s2)-1; /* length of s2 */

 if(s1==NULL || s2==NULL || n1<n2) /* when error exit */

   return 1;

}

int num_substring(void)

{

 int i, j, k;

 int count;

 for(i=0; i<=(n1-n2); i++) {

   count=0;

   for(j=i, k=0; k<n2; j++, k++){ /* search for the next string of size of n2 */

   if((s1+j)!=(s2+k)) {

     break;

   }

   else

     count++;

   if(count==n2)

     total++; /* find a substring in this step */

   }

 }

 return total;

}

int main(int argc, char argv[])

{

 int count;

 readf(fp);

 count=num_substring();

 printf("The number of substrings is: %d\n", count);

 return 1;

}

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