I want to check if a number is prime. Here is my code:
#include <iostream>
using namespace std;
int main(){
int num;
int i, k = 0;
cin >> num;
for(i = 2; i < num; i++){
if(num % i == 0){
k = k + 1;
}
}
if(k > 0){
cout << "The number is not prime" << endl;
}else{
cout << "Prime!" << endl;
}
return 0;
}
When I enter 6 , 78, ... etc it's giving the correct output.
But when I'm entering 4294967296 which is not a prime number, it's returning Prime!.
That's because 4294967296 is a rather special number. It is 2^32, which is limit of the storage capacity of an int, which typically holds only 32 bits. Thus, your num is interpreted as if it were 0.
The rest follows logically from that overflow misinterpretation. The for loop is never entered, since 2 is not less than 0, so k never gets incremented.
EDIT Actually, in this particular case, it fails for a different reason. std::cin performs truncation on the input, so when you enter 4294967296, num will actually get assigned the value of 2147483647. Your program [correctly] prints that that value is prime. It's just that that's not the number the user had intended to test.
You need to validate your input. 4294967296 is too big for 32 bit signed integer, so std::cin truncates it to 2147483647 which is actually prime. You can check that printing your number back to screen. You should also change your loop to:
for(i = 2; i < num / 2; i++)
as it would provide the same result, but decrease calculation time twice.
Related
This is my code for finding prime numbers between two integers. It compiles alright but giving a runtime error SIGXFSZ on codechef.
#include <bits/stdc++.h>
using namespace std;
int main() {
long long n,m;
int t;
cin>>t;
while(t--)
{
cin>>m>>n;
for(long long j=m;j<=n;j++)
for(long long i=2;i<=sqrt(j);i++)
if(j%i==0)
break;
else cout<<j<<"\n";
cout<<"\n";
}
return 0;
}
Seems that you are wrong on logic.
According to my understanding, you are supposed to print the prime numbers between two numbers.
But your code has logical errors.
1) Code doesn't consider 2 and 3 as prime numbers.
Say, m = 1, n = 10. For j = 2, 3, the inner loop won't execute even for the single time. Hence, the output won't be shown to be user.
2) else cout<<j<<"\n"; statement is placed incorrectly as it will lead to prime numbers getting printed multiple times and some composite numbers also.
Example:
For j = 11, this code will print 11 twice (for i = 2, 3).
For j = 15, this code will print 15 once (for i = 2) though it is a composite number.
You've underexplained your problem and underwritten your code. Your program takes two separate inputs: first, the number of trials to perform; second, two numbers indicating the start and stop of an individual trial.
Your code logic is incorrect and incomplete. If you were to use braces consistently, this might be clear. The innermost loop needs to fail on non- prime but only it's failure to break signals a prime, so there can't be one unless the loop completes. The location where you declare a prime is incorrect. To properly deal with this situation requires some sort of flag variable or other fix to emulate labelled loops:
int main() {
int trials;
cin >> trials;
while (trials--)
{
long long start, stop;
cin >> start >> stop;
for (long long number = start; number <= stop; number++)
{
if (number < 2 || (number % 2 == 0 && number != 2))
{
continue;
}
bool prime = true;
for (long long odd = 3; odd * odd <= number; odd += 2)
{
if (number % odd == 0)
{
prime = false;
break;
}
}
if (prime)
{
cout << number << "\n";
}
}
}
return 0;
}
The code takes the approach that it's simplest to deal with even numbers and two as a special case and focus on looping over the odd numbers.
This is basically "exceeded file size", which means that the output file is having size larger than the allowed size.
Please do check the output file size of your program.
I'm trying to find all the prime numbers between two integers and place them in an integer array.
The catch is that i have to use a specific method of doing so (divide each subsequent integer by all the primes in my array). So I can't use the sieve of Eratosthanes or any other 'easier' methods.
My code successfully prompts the user for two integers, but for now I do not use either of them. First I want to make sure the program works for values between 0 and whatever, in this case 200 just to test it.
Problem is, when I run the program and print the first 20 or so values in the array, I'm getting
2, 3, 5, 7, 11, 200, 0, 0, 0, 0, 0, 0 ...... more zeroes.
The first 5 values are correct because they start in the array, but after that the whole thing goes haywire.
I've worked through my nested loop by hand for a couple values and it SEEMS like it should work. I feel like there's a specific array property that I'm overlooking.
Here's my code:
#include "stdafx.h"
#include "iostream"
#include "climits"
#include "cmath"
#include "array"
using namespace std;
int main()
{
// declare variables to store user input
int lowerBound, upperBound;
// prompt user for lesser and greater integers and store them
cout << "Program to find all primes between two integers." << endl;
cout << "Enter lesser integer: " << endl;
cin >> lowerBound;
cout << "Enter greater integer: " << endl;
cin >> upperBound;
// if statement to switch the input variables if the user accidentally enters them backwards
if (lowerBound > upperBound) {
int temp = lowerBound;
lowerBound = upperBound;
upperBound = temp;
}
// initialize int array with the first 5 primes
int primes[100] = { 2, 3, 5, 7, 11 };
// loop to find primes between 12 and 200 (since we already have primes from 1-11 in the array)
for (int i = 12; i <= 200; i++) {
// the maximum divisor needed to determine if the current integer being tested is prime
double maxDivisor = sqrt(i);
// variable for the current size of the array
int size = 5;
// boolean variable is set to true by default
bool isPrime = true;
for (int j = 0; j < size; j++) { // changed "j<=size" to "j<size"
int remainder = (i % primes[j]);
// once the maximum divisor is reached, there is no need to continue testing for the current integer
if (primes[j] > maxDivisor) {
break;
}
// if the remainder of divison by a prime is 0, the number is not prime, so set the boolean variable to false
if (remainder = 0) {
isPrime = false;
}
}
// if isPrime is still true after the nested loop, the integer value being tested will be placed in the next element of the array
if (isPrime == true) {
primes[size] = i;
// since we added to the array, increment size by 1
size++;
}
}
// display the first 20 values in the array for debugging
for (int k = 0; k < 20; k++) {
cout << primes[k] << ", ";
}
system("pause");
return 0;
}
This here
if (remainder = 0) {
isPrime = false;
}
Needs to be changed to
if (remainder == 0) {
isPrime = false;
}
Because = does assignment, not comparison. So what remainder = 0 does it setting remainder to 0, and then it returns that 0, which gets casted to false, which is on of the reasons why it's not finding any primes.
Also, as Fantastic Mr Fox pointed out, for (int j = 0; j <= size; j++) needs to be changed to for (int j = 0; j < size; j++).
Also, did your compiler issue any warnings? If not, try to see if you can set it to be more strict with warnings. I figure most modern compilers will give you a hint at if (remainder = 0). Getting useful warnings from the compiler helps a lot with preventing bugs.
Edit:
As Karsten Koop pointed out, you need to move the int size = 5; out of the loop, to before the for (int i = 12;. With those changes, it's now working on my machine.
Last but not least, a tip: instead of if (isPrime == true), you can just write if (isPrime).
Closed. This question needs debugging details. It is not currently accepting answers.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
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The program I've written is supposed to take in two user inputs (one being the number we're meant to check whether it's k-hyperperfect or not, the other being a maximum k-value.) if the input integer is k-hyperperfect in the range of 1 to the inputted maximum k-value, then the output should be that k-value. For example, if the input integer is 21 and the maximum k-value is 100 then the output should be 2.
My program gives the correct output for (the first number is the input integer, the second number is the k-max value, the third number is output value) ...
21 (input integer) 100 (k-max) --> 180
301 100 --> 6
12211188308281 100 --> 0
-301 100 --> 0
21 -5 --> 0
However, it doesn't correctly execute for 12211188308281 and 200 (it gives me 0 when it should give me 180). I've run my code through a step by step visualizer and it seems to just abruptly stop execution when i = 496 in the for loop within the else statement. But I don't understand why since it executes correctly for 5 other test runs.
#include <iostream>
using std::cout; using std::cin; using std::endl; using std::fixed;
int main () {
int number;
int kmax;
int sum = 0 ;
int hyper = 0;
std::cin >> number;
std::cin >> kmax;
if (number <= 6 or kmax < 1) {
std::cout << "0" << "\n";
}
else {
for (int i=1;i<=number;i++) {
if (number%i==0 and i != 1 and i != number){
sum+= i;
}
}
}
for (int k=1; k <= kmax; k++) {
hyper = ((sum)*k) + 1;
if (hyper == number) {
std::cout << k << endl;
break;
}
}
}
You need to check that numbers read through std::istreams (like std::cin) are read successfully. As the value that you enter for number is too large to store in an integer your read will fail. For example you could change your code to:
int main()
{
int number;
std::cin >> number;
if ( !std::cin )
{
std::cout << "invalid value: " << number << "\n";
return 1;
}
else
{
std::cout << "valid value: " << number << "\n";
}
// calculate answer
return 0;
}
You would then see your program printing "invalid value: 2147483647" if you have a c++11 compliant compiler or an undefined number if you have an older compiler.
Now that you have implemented reading values correctly the fix to your issue is to use a larger integer type like int64_t which is able to hold your number.
As already noted, the int type in your machine isn't big enough to store the value 12,211,188,308,281.
The C++ standard only mandates it to be capable of storing a value up to 32,767 and even in the (now common) case of a 32-bit int or long int), the limit would be 2,147,483,647. So you need a long long int or an int64_t (if it's present in your implementation).
A simple check like
if (std::cin >> number >> kmax ) { // perform calculations...
Would have shown the error sooner.
That beeing said, there are also some simple changes that could be done to the posted code in order to make it more efficient. The first loop can be optimized considering the "symmetry" of the divisors of a given number: meaning, if n is divisible by a, so that b = n/a is a whole number, b too is a divisor of n. This will limit the number of iterations to the square root of n, instead of n.
long long int number,
kmax,
sum = 0;
// ...
long long int temp = number,
i = 2;
for (; i * i < number; i++) {
if (number % i == 0) {
temp = number / i;
sum += i + temp;
}
}
if (i * i == number) {
sum += i;
}
There probably are better algorithms, but I'm unfamiliar with those.
The second loop, in my opinion, is unnecessary. The value k can be calculated directly:
if ( (number - 1) % sum == 0) {
std::cout << (number - 1) / sum << '\n';
}
You are assigning a too long value 12211188308281 to integer "number", which can't contain it fully and it is getting truncated to 596285753. You can add a print statement to print it.
std::cout<<number;
which will print 596285753.
As suggested you should use long long int. Again its dependent on the software platform running on your system.
I'm writing a program for finding whether a given number is an Armstrong Number:
int main()
{
int n,rem,sum=0;
cout<<"Enter the Number for checking"<<endl;
cin>>n;
while(n!=0)
{
rem=n%10;
sum=sum+(rem*rem*rem);
n=n/10;
}
if(sum==n)
{
cout<<"Armstrong Number"<<endl;
}
else
{
cout<<"It's not a armstrong number";
}
return 0;
}
When I run it, it always reports "It's not a armstrong number", regardless of input.
I changed the code as follows, and got the correct result. But I don't understand why I need to assign input to n1 and do the operation - why can't I directly do the operation with n?
int main()
{
int n,rem,sum=0,n1;
cout<<"Enter the Number for checking"<<endl;
cin>>n;
n1=n;
while(n1!=0)
{
rem=n1%10;
sum=sum+(rem*rem*rem);
n1=n1/10;
}
if(sum==n)
{
cout<<"Armstrong Number"<<endl;
}
else
{
cout<<"It's not a armstrong number";
}
return 0;
}
In the line if (sum==n) your program compares sum and n. In the second program n is initial number entered by user. But in the first program n==0 (see the loop above it).
So, in the first program the check if (sum==n) works as if (sum==0). But value of sum is never 0 (except user entered 0). So, first program always returns "It's not a armstrong number".
And about style: It is much better to use functions instead of putting the whole logic into one main() function. For instance, you can create a function for calculation of the intermediate sum for cheching of Armstrong Number:
int getSumOfCubesOfDigits(int n)
{
int sum = 0;
while (n)
{
const int rem = n % 10;
sum += rem * rem * rem;
n = n / 10;
}
}
In this case your program will be much simpler and it will be hard to make the mistake you have in the first program of your question:
int main()
{
int n;
cout << "Enter the Number for checking" << endl;
cin >> n;
if(getSumOfCubesOfDigits(n) == n)
cout<<"Armstrong Number"<<endl;
else
cout<<"It's not a armstrong number";
return 0;
}
In the first program, the original number is entered into 'n'. The only problem in your logic is, you forgot that by the time you exit from the while loop, 'n' will no longer be your original number since you are repeatedly doing n=n/10, and hence 'sum==n' never satisfies even for an Armstrong number.
So before you enter the while loop, save the original number into another variable, say n1 (as done in the second program you provided), and only use n1 for operations, ie, n1=n1/10. Leave n alone so that, in the end 'n' will still contain the original number, which you can finally compare with 'sum' to find your answer.
Which number do you compare ? , in first program in while loop , n value is changed ( in this variable you get the input) and finally check with sum == n , so it always get condition fail.
So temp (n1) variable is required , to compare the final result
Your code is different in the second code block, you are still testing if sum=n.
In the second code block, if you tested if(sum=n1), I would suspect it would work the same.
I got this solution for finding an Armstrong Numbers
int main() {
for (int i=10; i<=9999; i++) {
int k = i,z = 1, s = 0, n = i;
while ((k/=10) > 0) z++;
for (int t = z; t; t--, n/=10) {
s += pow(n % 10, z);
}
if (i == s) cout << i << endl;
}
return 0;
}
I am working on a program in which I must print out the number of primes, including 1 and 239, from 1 - 239 ( I know one and or two may not be prime numbers, but we will consider them as such for this program) It must be a pretty simple program because we have only gone over some basics. So far my code is as such, which seems like decent logical flow to me, but doesnt produce output.
#include <iostream>
using namespace std;
int main()
{
int x;
int n = 1;
int y = 1;
int i = 0;
while (n<=239)
{x = n % y;
if (x = 0)
i++;
if (y < n)
y++;
n++;
while (i == 2)
cout << n;
}
return 0;
}
The way I want this to work is to take n, as long as n is 239 or less, and preform modulus division with every number from 1 leading up to n. Every time a number y goes evenly into n, a counter will be increased by 1. if the counter is equal to 2, then the number is prime and we print it to the screen. Any help would be so greatly appreciated. Thanks
std::cout << std::to_string(2) << std::endl;
for (unsigned int i = 3; i<240; i += 2) {
unsigned int j = 3;
int sq = sqrt(i);
for (; j <= sq; j += 2) if (!(i%j)) break;
if (j>sq) std::cout << std::to_string(i) << std::endl;
}
first of all, the prime definition: A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself.
so you can skip all the even numbers (and hence ... i+=2).
Moreover no point to try to divide for a number greater than sqrt(i), because then it will have a divisor less than sqrt(i) and the code finds that and move to the next number.
Considering only odd numbers, means that we can skip even numbers as divisors (hence ... j+=2).
In your code there are clearly beginner errors, like (x = 0) instead of x==0. but also the logic doesn't convince. I agree with #NathanOliver, you need to learn to use a debugger to find all the errors. For the rest, good luck with the studies.
lets start with common errors:
first you want to take input from user using cin
cin>>n; // write it before starting your while loop
then,
if (x = 0)
should be:
if (x == 0)
change your second while loop to:
while (i == 2){
cout << n;
i++;
}