1. Using vectors or arrays, write a function named word_rev() that reverse any g

Question

1. Using vectors or arrays, write a function named word_rev() that reverse any given word.

2. Write a program that will:

Ask the user to enter a word.

Save the entry as word_entered.

Call word_rev on the entry

Display the reversed entry (word_entered)

3. Write a function named prime() that determine whether or not a given number n (greater than

one) is prime. The algorithm:

If n is even then n is not a prime number except n = 2

For all odd numbers less or equal to the square root of n, if any of them divide n, then

n is not prime

4. Write a program that:

Asks the user to enter an integer

Saves the entry as var

Calls the function prime() on the entry

Displays whether or not the entry is prime

5. Write a function named eratos() that find all the prime number less than a given number, its

argument n

6. Write a program that:

Asks the user to enter an integer

Saves the entry as bound

Passes bound onto eratos()

Displays all the prime numbers less or equal to bound

## the code does not build

Explanation / Answer

1.
void word_REV(string& y)
{
string temp = "";
int len = y.length();

for(int i = len - 1; i >= 0; --i)
{
temp += y[i];
}
y = temp;
}

2.

#include <iostream>
#include <string>

using namespace std;
void word_REV(string& y)
{
string temp = "";
int len = y.length();

for(int i = len - 1; i >= 0; --i)
{
temp += y[i];
}
y = temp;
}

int main()
{
string word_entered;
cout<< "enter the word to be reversed";

cin>>word_entered

reverseStr(word_entered);

cout << "word_entered = " << x << endl;

return 0;
}

3.

int prime(double num)
{
   int isprime = 0;
   for(int i = 2; i <= sqrt(num); i += 2)
   {
       if(i % 2 == 0)
           i++;

       if((int(num)% i) == 0)
       {
           isprime = 1;
           break;
       }
   }

   return isprime;
}

4.
#include <math.h>
#include <iostream>

using namespace std;
using namespace System;

int prime(double num);
int prime(double num)
{
   int isprime = 0;
   for(int i = 2; i <= sqrt(num); i += 2)
   {
       if(i % 2 == 0)
           i++;

       if((int(num)% i) == 0)
       {
           isprime = 1;
           break;
       }
   }

   return isprime;
}
int main()
{
double var;
cin>>"enter number to determine";
cin>>var;
if(prime(var)==0)
cout>>"given integer is prime";
cout>>"given integer is not prime";
return(0);
}

5.

void eratos(int n)
{
bool prime[n+1];
memset(prime, true, sizeof(prime));

for (int p=2; p*p<=n; p++)
{
// If prime[p] is not changed, then it is a prime
if (prime[p] == true)
{
// Update all multiples of p
for (int i=p*2; i<=n; i += p)
prime[i] = false;
}
}

// Print all prime numbers
for (int p=2; p<=n; p++)
if (prime[p])
cout << p << " ";
}

6.

#include <bits/stdc++.h>
using namespace std;

void eratos(int n)
{
bool prime[n+1];
memset(prime, true, sizeof(prime));

for (int p=2; p*p<=n; p++)
{
// If prime[p] is not changed, then it is a prime
if (prime[p] == true)
{
// Update all multiples of p
for (int i=p*2; i<=n; i += p)
prime[i] = false;
}
}

// Print all prime numbers
for (int p=2; p<=n; p++)
if (prime[p])
cout << p << " ";
}

int main()
{
int bound;
cout<< "enter number ";
cin>>bound;
cout << "Following are the prime numbers smaller "
<< " than or equal to " << n << endl;
eratos(n);
return 0;
}

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