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Nina [5.8K]
3 years ago
12

PLZZZ HELP!!! I’ll give brainliest

Computers and Technology
1 answer:
prohojiy [21]3 years ago
3 0

Answer:

The ability of a firm to improve contribution margins (revenue-variable costs) as its sales volume increases. Infinite scalability can occur only when the variable cost of serving an additional customer is zero.

Explanation:

The best that I can do

Hope this helps : )

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A transmission system at a radio station uses a/an _______ to convert a direct current into a high-frequency alternating current
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network service providers

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Which of the following statements about Java Class Libraries is false? Question 18 options: A) An advantage of using Java class
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D) Java class libraries are not portable

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Can someone who is good at 9th class IT (cbsce) can help me in my homework​
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Answer: yeah

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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;

   do

   {

       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
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