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I'm currently trying to detect and print prime numbers. Here's what I've got for now, but I get an error from the compiler: non-lvalue in assignment, and I can't seem to be able to fix that. Any help?
Code:#include<iostream>
using namespace std;
int main() {
int i, j;
for(i=0; i<100; i++) {
for(j=0; j<i; j++) {
if(i%j=1 && i%j=i) {
cout<<"Prime"<<endl;
}
else {
cout<<"Not Prime"<<endl;
}
}
}
}
should beCode:if(i%j=1 && i%j=i) {
= for assignment, == for equality. :thumb:Code:if(i%j==1 && i%j==i) {
You have also ill-formed logic for your task. Consider, that % operator returns the remainder, not the bool value 'divisable'. For determining whether i is divided by j, you have to determine whether the i%j value is zero, and vice versa.
Also note:
Code:anything%0; //divide by zero error
agree with amadeus, use % to detect non-primes if its equal to 0 for any j, if not 0 for any j>1 then say its prime
another way is
thus at the end nonprime array is a bitvector, if a index is 1 its nonprime (i.e. nonprime[50]=1 while nonprime[47]=0 at the end of execution )Code:int nonprime[100];//initialize all elements to 0
for(i=0; i<100; i++) {
for(j=2; j<=i; j++) {
nonprime[i*j]=1;
}
I re-wrote the program, and it only works for interval 1-3. The problem is within the second loop, I guess.
Code:#include<iostream>
using namespace std;
int main() {
int l;
cout<<"Please enter an interval limit (0-N): ";
cin>>l;
for(int i=0; i<=l; i++)
for(int n=1; n<l; n++)
if(i%n==1)
cout<<i<<" ";
return 0;
}
Again, i % n == 1 is a check whether i is divided by n with remainder 1. That is not probably what you want. If you want to check whether i has a divider different from i and 1, then you should (in your way) iterate all the numbers from 2 to (i - 1), and if somewhere i % n == 0 (say n is the iterator), then i is not prime. Otherwise, if the loop reaches its end, the i is prime.
Note, that there are lots of algorithms for detecting primes, and iterating from 2 to (n - 1) is not neccessary, since it takes much time on presumably unneccessary checks (you could increase step to 2, iterate up to sqrt(n - 1), etc) - but the one you are trying is the simplest one for understanding your task.
Hope that helps.
That's also something I've tried. The original program had n=2; n<(i-1). Here's the latest version. It should work (if you enter 3, then it checks if 1:2, 2:2 or 3:2 has remainder 0, and if it doesn't, display the number. it displays me no numbers...).
Code:#include<iostream>
using namespace std;
int main() {
int l;
cout<<"Please enter an interval limit (0-N): ";
cin>>l;
for(int i=1; i<=l; i++)
for(int n=2; n<=(i-1); n++)
if(i%n!=0 && i%n==i && i%n==1)
cout<<i<<endl;
return 0;
}
Not surprisingly that no numbers are displayed. Look at your condition - it is always false since i%n cannot be both i and 1 at the same time. Why do you still code that annoying i%n==1 check? What are going to see? Again - i%n - is the REMAINDER, and i%n==1 means that i = n*k + 1... And: i%n==i - is totally impossible when i >= n, since the remainder is 0..(n-1) number. So, throw those two annoying bad checks away - and invert logic of the first one to i%n==0 - then your program will display non-prime numbers.
Ok, now it does print non-prime numbers, but it displays each of them too many times. The output for 10 is something like 4 6 6 8 8 9 10 10. I can figure out I have to make it pass to the next value as soon as i checks (for example, if 6%2 is true, don't check 6%3, but get to 7%2).
I know I'm doing it the wrong way, but I'm trying to actually display PRIME numbers. Using the non-primes algorithms, I'd like to create a vector of l elements, including integers from 1 to l. After using the algorithm, erase the generated numbers from the vector, and then print the rest of the numbers. I'm excluding the non-prime elements from the interval. But can I erase without specifying a position?
i write the code for you:
according to a math theory it is just enough to test for 0 remainder till reaching sqrt(i). so i used that theory to have faster algorithmCode:#include <cstdlib>
#include<iostream>
#include<math.h>
using namespace std;
int main() {
int i, j;
for(i=2; i<100; i++)
{
bool isPrime = true;
for(j=2; j<=sqrt(i); j++)
{
if(i%j ==0 )
{
isPrime = false;
break;
}
}
if (isPrime)
{
cout<< i << " : Is Prime" << endl;
}else
{
cout<< i << " : Is not Prime" << endl;
}
}
system("PAUSE");
return EXIT_SUCCESS;
}
I appreciate this very much, but I'm not necessarily trying to print prime numbers. I'm reading a book, and the exercise wants me to detect and print prime numbers using 2 for loops and the modulus operator.
Here's my latest code. I put that break over there because w/o it, the program would have checked for numbers which would not eventually give the remainder 0. Now it works fine, but it does print the ones which are divided by odd numbers (for example, on 1-16, it prints 9 and 15, but excludes any number which is dividable by 2). I know n doesn't go further than 2 because of that break, and it drives me crazy.
Code:#include<iostream>
using namespace std;
int main() {
int l;
cout<<"Please enter an interval limit: ";
cin>>l;
for(int i=1; i<=l; i++) {
for(int n=2; n<l; n++) {
if(i%n!=0) {
cout<<i<<endl;
}
break;
}
}
return 0;
}
Though if you actually want to benefit from the theory, you should compute the square root just once, store the result, and then use that in the loop.Quote:
Originally Posted by toraj58
EDIT:
s21122012, you might want to consider naming your variable with a better name than l. For one thing, a one letter variable name is not descriptive in that context, and for another thing, l looks like 1 on some fonts, including the one this message board uses by default. You also need to indent your code better.
yes that was also a performance hit.
it should be out of the second loop.
Furthermore, you can replace ++i in both loops with i+=2 with some slight check adding outside the loop.
Ok, guys. That's some stuff I haven't heard about yet. Getting back to my problem, the following code (I hope it's better now, at least esthetically) generates the following output. It does not display the same numbers a billion times, but it skips all the divisors except 2, which is the initial value of N. So the second loop is not working, most probably because of the break keyword, but without it, I get all the numbers from 2 to limit.
OUTPUT:Code:#include<iostream>
using namespace std;
int main() {
int limit;
cout<<"Please enter an interval limit: ";
cin>>limit;
for(int i=1; i<=limit; i+=2) {
for(int n=2; n<limit; n+=2) {
if(i%n!=0) {
cout<<i<<" ";
}
break;
}
}
return 0;
}
Code:Please enter an interval limit: 50
1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49
Process returned 0 (0x0) execution time : 2.453 s
Press any key to continue.
it is wrong:
9, 15,21 and ... are not prime.
#include "stdafx.h"
#include <iostream>
using namespace std;
int main()
{
int i, j;
for ( i = 2; i < 100; i++ ) {
for ( j = 2; j <= i/2; j++ ) {
if ( ! ( i % j ) ) break;
}
if ( j > i / 2 ) cout << i << endl;
}
return 0;
}
This code works great, but can someone help explain this loop to me? I'm especially having a hard time with the if ( ! ( i % j ) ) break; statement.
This is a (not very) shorthand way of sayingCode:if ( ! ( i % j ) )
It relies on a non-zero being interpreted as 'true' and zero as 'false'. The '!' inverts the logic.Code:if ((i % j) == 0)
Therefore, if the value of i % j is zero, then that is interpreted as 'false'. the '!' inverts this to 'true'.
I find the second example preferable as it states the test explicitly.
just to add more verboseness, are you aware that false is '0' and true is 'not zero'?
so if x%y == 0, then that is equivalent (with respect to bools) to x%y == false, otherwise it is true.
just fyi.
Hey thanks Amleto for the help, it took me a little while to wrap my brain around the loop sequence, and after I put in cout << "i = " << i << " j = " << j << endl; after the second loop, I could see what it was doing with i and j.
I know this problem has been solved, but for anyone else who is interested. I has been shown that all prime numbers reside around multiples of 6. So you could slightly speed up the app by adding +=6 instead of one or two. See the code below.
Code:#include <vector>
#include <iostream>
#include <iomanip>
using namespace std;
// playing with primes
int main() {
vector<int> actualPrimes;
actualPrimes.push_back(2);
actualPrimes.push_back(3);
for( int i=6; i<1093/*arbitrary*/; i+=6 ) {
bool left=true,right=true;
int endSize = actualPrimes.size(),
endDelim = i/2;
for(int j=0; j<endSize && actualPrimes[j]<endDelim; ++j) {
if( (i-1) % actualPrimes[j] == 0 ) {
left = false;
} if( (i+1) % actualPrimes[j] == 0 ){
right = false;
}
}
if( left ) {
actualPrimes.push_back(i-1);
}if(right) {
actualPrimes.push_back(i+1);
}
}
cout<<"The list of primes are:";
for( int i=0; i<actualPrimes.size(); ++i) {
if( i%6 == 0 ) cout<<endl;
cout<<setw(10)<<actualPrimes[i]<<" ";
}
return 0;
}
I was interested in this, so I spat out the code in 5 minutes.
All the best!