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xenn [34]
3 years ago
7

Assume the availability of a function is_prime. Assume a variable n has been associated with positive integer. Write the stateme

nts needed to find out how many prime numbers (starting with 2 and going in increasing order with successively higher primes [2,3,5,7,11,13,...]) can be added before exceeding n. Associate this number with the variable k. SUBMIT
Computers and Technology
1 answer:
My name is Ann [436]3 years ago
5 0

Answer:

def main():

   # Accept data from the user

   n=int(input('Enter the value of n:'))

   k=2;

   sum=0

   print('The list of the prime numbers are as follows:')

   # This loop runs from k to n

   while k<=n:

       # Call is_prime() method,

       # to check whether k is a prime or not

       sum=sum+is_prime(k)

       k=k+1

   print('Sum of the prime numbers:',sum)

# is_prime() Method

def is_prime(k):

   prime=0        

   i=1

   # Check k is a prime or not

   while i<=int(k):

       # Find the factors of k

       if (k % i) == 0:

           prime=prime+1

       i=i+1

   # If k as two factors

   if(prime==2):        

       print(k)

       return k;

   # If k as more than two factors

   else:        

       return 0;

# Call the main() method

main()

Explanation:

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Answer:

#include <iostream>

#include <string>

#include <cstring>

#include <cstdlib>

using namespace std;

struct Person

{

string name;

int age;

};

int main()

{

struct Person data[10];

struct Person *pData[10],*temp;

string names[] = {"a","b","c","g","z","l","p","q","r","w"};

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for(int i=0;i<9;i++)

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data[i].name = names[i];

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{

if(pData[i]->name.compare(pData[j]->name)>0)

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temp = pData[i];

pData[i] = pData[j];

pData[j] = temp;

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for(int i=0;i<9;i++)

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cout<<pData[i]->name<<" "<<pData[i]->age<<endl;

}

}

Explanation:

The line #include <iostream> initializes the program.

This program created an array of only 10 structures, Customer data. Also defined an auxiliary array Customer *pData.

The program uses these parameters to sorts the array of pointers so that when you go through pData in increasing order of index k, the entries pData[k] point to Customer objects in decreasing order by balance, meaning that pdata[0] now points to the customer with the highest balance, and pData[9] points to the customer with the smallest balance.

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