Why is this code not printing the prime factors of num? - c++

I wrote this code for obtaining the prime factors of a number taken as an input from the user.
#include<bits/stdc++.h>
using namespace std;
void prime_Factors(int);
bool isPrime(int);
int main()
{
int num;
cout << "Enter the number to find it's prime factors: ";
cin >> num;
prime_Factors(num);
}
void prime_Factors(int n1)
{
for(int i = 2; i<n1; i++)
{
if(isPrime(i))
{
int x = i;
while(n1%x==0)
{
cout << i << " ";
x *= i;
}
}
}
}
bool isPrime(int n0)
{
if(n0==1)
return false;
for(int i = 0; i*i <= n0; i++)
{
if(n0%i==0)
return false;
}
return true;
}
The prime_Factors() function call in main() function is not printing the prime factors. Pls help!!

The ranges of the loops are wrong.
Firstly, the loop for(int i = 2; i<n1; i++) will fail to find prime factors of prime numbers (the numbers theirself). It should be for(int i = 2; i<=n1; i++).
Secondly, the loop for(int i = 0; i*i <= n0; i++) will result in division-by-zero. It should be for(int i = 2; i*i <= n0; i++).

Thinking about using the Sieve of Eratosthenes made me try it out:
#include <iostream>
#include <cstdint>
#include <vector>
void prime_factors(uint32_t n) {
while(n % 2 == 0) {
std::cout << "2 ";
n /= 2;
}
std::vector<bool> sieve(n / 2, true);
for (uint32_t i = 3; i * i <= n; i += 2) {
if (sieve.at(i / 2 - 1)) {
uint32_t j = i * i;
for (; j < n; j += 2 * i) {
sieve.at(j / 2 - 1) = false;
}
if (j == n) {
do {
std::cout << i << " ";
n /= i;
} while (!sieve.at(n / 2 - 1));
}
}
}
if (n > 1) std::cout << n;
std::cout << "\n";
}
int main() {
prime_factors(123456789);
}
https://godbolt.org/z/8doWbYrs6

Related

print all the prime numbers upto an upperlimit

#include<iostream>
using namespace std;
int main(){
int n=5;
int i = 2;
for (i; i <= n; i++)
// for all num to n
{
int j = 2;
bool divide = false;
for (j; j <= n - 1; j++)
// for checking each num
{
if (i % j == 0)
{
divide = true;
break;
}
}
if (divide == false)
{
cout << i << " ";
}
}
return 0;
}
my Q is that
//please tell me why it is not working
//it is expected to give ans 2,3,5 which it is not giving why???
maybe I found the issue.
I think that the problem here is:
for (j; j <= n - 1; j++)
Here you did j<=n-1;
So to fix this just do:
for(j; j < i; j++){
//this should fix
So everything should look like this:
#include<iostream>
using namespace std;
int main() {
int n = 5;
int i = 2;
//check prime numbers starting from i and max n using for loop
for (i = 2; i <= n; i++) {
bool divide = false;
for (int j = 2; j < i; j++) {
if (i % j == 0) {
divide = true;
break;
}
}
if (!divide) {
//!divide is equal to divide=false
cout << i << " ";
}
}
}

What's wrong with factorial function code in this Pascal triangle?

can anyone explain why first fact() function code doesn't give me correct output but second one does?
what's wrong with the commented factorial function code???
#include <iostream>
using namespace std;
// int fact(int n){
// for (int i = n-1; i > 0; i--){
// n = n * i;
// }
// return n;
// }
int fact(int n){
int temp=1;
for (int i = 2; i <=n; i++){
temp= temp * i;
}
return temp;
}
int nCr(int n, int r){
int temp=(fact(n) / (fact(n - r) * fact(r)));
return temp;
}
int main(){
int n;
cout << "enter no: ";
cin >> n;
for (int i = 0; i < n;i++){
for (int k = 1; k < (n - i);k++)
cout << " ";
for (int j = 0; j <= i; j++)
{
cout << nCr(i, j) << " ";
}
cout << endl;
}
return 0;
}
The second fact function correctly returns 1 when n = 0, but the first one wrongly returns 0 when n = 0.
Adding check for this case will make the first function work well.
int fact(int n){
if (n == 0) return 1; // check for n = 0 case
for (int i = n-1; i > 0; i--){
n = n * i;
}
return n;
}

prime seive algorithm giving a runtime error

I am trying to implement the Sieve of Eratosthenes algorithm but it giving a runtime error.
didn't get any output though. after providing the input,
#include<iostream>
using namespace std;
//Sieve Approach - Generate an array containing prime Numbers
void prime_sieve(int *p) {
//first mark all odd number's prime
for (int i = 3; i <= 10000; i += 2) {
p[i] = 1;
}
// Sieve
for (long long int i = 3; i <= 10000; i += 2) {
//if the current number is not marked (it is prime)
if (p[i] == 1) {
//mark all the multiples of i as not prime
for (long long int j = i * i; j <= 10000; j = j + i ) {
p[j] = 0;
}
}
}
//special case
p[2] = 1;
p[1] = p[0] = 0;
}
int main() {
int n;
cin >> n;
int p[10000] = {0};
prime_sieve(p);
//lets print primes upto range n
for (int i = 0; i <= n; i++) {
if (p[i] == 1) {
cout << i << " ";
}
}
return 0;
}
compiler didn't throwing any error also it is not providing the output also
program freezes for some seconds and then terminates
As mentioned in the comments, you are going out of bound.
There is also some confusion about the meaning of p[].
In addition, you are not using the value of n in the function, which leads to unnecessary calculations.
Here is a tested programme (up to n = 10000):
#include <iostream>
#include <vector>
#include <cmath>
//Sieve Approach - Generate an array containing prime Numbers less than n
void prime_sieve(std::vector<int> &p, long long int n) {
//first mark all odd number's prime
for (long long int i = 4; i <= n; i += 2) {
p[i] = 0;
}
// Sieve
for (long long int i = 3; i <= sqrt(n); i += 2) {
//if the current number is not marked (it is prime)
if (p[i] == 1) {
//mark all the multiples of i as not prime
for (long long int j = i * i; j <= n; j = j + i ) {
p[j] = 0;
}
}
}
//special cases
p[1] = p[0] = 0;
}
int main() {
long long int n;
std::cout << "Enter n: ";
std::cin >> n;
std::vector<int> p (n+1, 1);
prime_sieve(p, n);
//lets print primes upto range n
for (long long int i = 0; i <= n; i++) {
if (p[i] == 1) {
std::cout << i << " ";
}
}
return 0;
}

How do I output all numbers with a given amount of divisors? (C++)

I need to enter 2 integers, n and k, n is the range and k is the amount of divisors for those numbers, and only display the numbers with k divisors.
#include <iostream>
using namespace std;
int main()
{
int n, k,cnt=0;
cin>>n;
cin>>k;
for(int i=1; i<=n; i++)
{
if(n%i==0)
{
cnt++;
}
if(k==cnt)
cout<<i<<" ";
}
return 0;
}
#include <bits/stdc++.h>
using namespace std;
int main(){
int n, k;
cin>>n>>k;
for(int i = 1; i <= n; i++){
int num_divisors_i = 0;
for (int j = 1, len = sqrt(i); j <= len; j++) {
if (i % j == 0) {
if (i / j == j) {
num_divisors_i++;
}
else {
num_divisors_i = num_divisors_i + 2;
}
}
}
if(num_divisors_i == k) cout<<i<<" has "<<k<<" divisors"<<endl;
}
}
This was my solution:
#include<iostream>
int main(int argc, char* argv[]) {
int n = 0, k = 0;
std::cout << "Enter the range: ";
std::cin >> n; if (!std::cin) throw std::runtime_error("range read failed");
std::cout << std::endl << "Enter the number divisors: ";
std::cin >> k; if (!std::cin) throw std::runtime_error("number of divisors read failed");
std::cout << "numbers with " << k << " divisor:: ";
for (int i = 1; i != n+1/*if k is inclusive else n*/; ++i) {
int cnt = 0;
for (int j = 1; j != i+1; ++j) {
if (i % j == 0)
cnt += 1;
}
if(cnt == k)
std::cout << i << ' ';
}
return 0;
}

Improve on binary converting algorithm (include negative numbers)

I'm doing some C++ array homework. The goals is to convert decimal to binary (include negative numbers). Here's my code, it gets the job done, but I would like to see if anything can be improved, or any better algorithm (using binary shift maybe?).
#include <iostream>
using namespace std;
// doi tu thap phan sang nhi phan
void decToBinary(int n, int nhiphan[])
{
for (int i=0; i < 16; i++)
{
// to binary
nhiphan[i] = n % 2;
n = n / 2;
}
// inverse array
for (int i = 0, j = 15; i < j; i++, j--)
{
int temp = nhiphan[i];
nhiphan[i] = nhiphan[j];
nhiphan[j] = temp;
}
}
void reverse(int& a)
{
if (a == 0)
a++;
else a--;
}
void outArr(const int a[], int size) {
for (int i = 0; i < size; ++i)
cout << a[i];
}
int main()
{
int nhiphan[16];
int n;
do {
cout << "Nhap so (-255 <= n <= 255) chuyen doi sang nhi phan (16 bit): ";
cin >> n;
} while (n > 255 || n < -255);
if (n < 0) {//check negative
n *= -1;
decToBinary(n, nhiphan);
for (int i = 0; i < 16; i++)// 1's complement
reverse(nhiphan[i]);
// +1
if (nhiphan[15] == 0)//2's complement
nhiphan[15] = 1;
else
{
nhiphan[15] = 0;
int i = 15;
do {
reverse(nhiphan[i-1]);
i--;
} while (nhiphan[i-1] == 0);
}
}
else decToBinary(n, nhiphan);
outArr(nhiphan, 16);
return 0;
}