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elixir [45]
4 years ago
15

Consider this data sequence: "3 11 5 5 5 2 4 6 6 7 3 -8". Any value that is the same as the immediately preceding value is consi

dered a CONSECUTIVE DUPLICATE. In this example, there are three such consecutive duplicates: the 2nd and 3rd 5s and the second 6. Note that the last 3 is not a consecutive duplicate because it was preceded by a 7.
Write some code that uses a loop to read such a sequence of non-negative integers , terminated by a negative number. When the code exits the loop it should print the number of consecutive duplicates encountered. In the above case, that value would be 3.
Computers and Technology
1 answer:
iris [78.8K]4 years ago
5 0

Answer:

Following is the source code required:

int entry_1 ,entry_2 = -1, consecutive_duplicates = 0;

do {

cin >> entry_1;

if ( entry_2 == -1)

{

entry_2 = entry_1;

}else

{

if ( entry_2  == entry_1 )

consecutive_duplicates++;

else

entry_2  = entry_1;

}

}

while(entry_1 > 0 );

cout << consecutive_duplicates;

Explanation:

Following is the explanation for given code:

  • 3 integers are declared as entry_1, entry_2  and consecutive_duplicates.
  • entry_2 = -1 (it will tell that the integers are non-zero)
  • user will enter the value for entry_1
  • The loop will be applied which will check for the value of entry_2. If entry_2 = -1, the value of entry_1 will be stored in entry_2 else it the value of entry_2 is already equal to entry_1, the integer consecutive_duplicate in increased by one.
  • Now the value of entry_1 is stored in entry_2 , so that next entity might be checked.
  • In the end, while the entry_1 is greater than zero (unless the last (negative)element of array is reached), print the integer consecutive_duplicates.

i hope it will help you!

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svp [43]

Answer:

69.08265412 milliseconds

Explanation:

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7*1024*1024*8=58720256 bits

Now convert bits to Gbits

58720256/10^{9}  =0.058720256 Gbits

Queuing Delay = Total size/transmission link rate

Queuing Delay= \frac{0.058720256}{1.7} =0.03454132706 seconds

Delay of packet number 3 = 0.03454132706*2=0.06908265412 seconds

or 0.06908265412= 69.08265412 milliseconds

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Explanation:

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Write a C program that counts the number of vowels in a word. The C program reads words from an input file and then stores in an
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Answer:

Explanation:

The following code is written in C and does what the question requires. It uses the input file reads it and outputs the number of vowels to the file called output.dat

#include <ctype.h>

#include <stdio.h>

#include <stdlib.h>

#include <string.h>

int str_count_in_chars(char *start, char *end, char *chars) {

   int count = 0;

   for (; start != end; count += !!strchr(chars, *(start++)));

   return count;

}

void str_swap_in_chars(size_t str_len, char **str, char *chars) {

   for (int front = 0, back = str_len - 1; front < back; front++) {

       if (strchr(chars, (*str)[front])) {

           for (; !strchr(chars, (*str)[back]); back--);

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           (*str)[front] = (*str)[back];

           (*str)[back--] = tmp;

       }

   }

}

char *file_to_str(FILE *fin) {

   int buf_len = 64;

   char buf[buf_len];

   char *str = malloc(buf_len);

   str[0] = '\0';

   for (int i = 1; fgets(buf, buf_len, fin); i++) {

       if (!(str = realloc(str, i * buf_len))) {

           fprintf(stderr, "%s:%d realloc failed\n", __FILE__, __LINE__);

           exit(1);

       }

       strcat(str, buf);

   }

   return str;

}

int main() {

   char *vowels = "aeiou";

   FILE *fin = fopen("input.dat", "r");

   FILE *fout = fopen("output.dat", "w");

   if (!fin || !fout) {

       fprintf(stderr, "%s:%d fopen failed\n", __FILE__, __LINE__);

       exit(1);

   }

   char *words = file_to_str(fin);

   fclose(fin);

   int words_len = strlen(words);

   for (int i = 0; i < words_len;) {

       if (isspace(words[i])) {

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       }

       int start = i;

       for (; i < words_len && !isspace(words[i]); i++);

       char *word = words + start;

       int word_len = i - start;

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   fclose(fout);  

   free(words);

   return 0;

}

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Answer:

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Explanation:

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