Ok, so I enjoy using SPOJ to practice programming and developing algorithms. I always have issues with the questions though. A lot of times, I will get a "wrong answer" message when clearly my code answers the questions properly. If someone could tell me if there is anything wrong or why SPOJ would be telling me my answer was wrong that would be awesome! Here is the problem word-for-word:
Prime Number Generator
Peter wants to generate some prime numbers for his cryptosystem. Help him! Your task is to generate all prime numbers between two given numbers!
Input
The input begins with the number t of test cases in a single line (t<=10). In each of the next t lines there are two numbers m and n (1 <= m <= n <= 1000000000, n-m<=100000) separated by a space.
Output
For every test case print all prime numbers p such that m <= p <= n, one number per line, test cases separated by an empty line.
My code:
int n;
scanf("%d", &n);
if(n > 10){ return 0; }
n = n*2;
int arr[n];
for(int i = 0; i < n; i++){ scanf("%d", &arr[i]); }
for(int i = 0; i < n; i += 2){
if(arr[i] >= 1 && arr[i] <= arr[i+1] && arr[i+1] <= 1000000000 && arr[i+1]-arr[i] <= 100000){
for(int j = arr[i]; j <= arr[i+1]; j++){
if(j % 2 == 1 || j == 2){
printf("%d\n", j);
}
}
printf("\n");
}
}
return 0;
INPUT:
2
7 11
2 9
OUTPUT:
7
9
11
2
3
5
7
9
A lot of times, I will get a "wrong answer" message when clearly my code answers the questions properly.
This is not one of those cases as evidenced by the fact that, despite the contrary, your code seems to think that 9 is a prime. The line:
if(j % 2 == 1 || j == 2)
combined with the fact that you appear to be printing all odd numbers (and two), is an indication that your prime check is incorrect.
Where you should probably start is with a simple prime check function such as:
int isPrime(int num) {
int chk = 2;
while (chk * chk <= num)
if ((num % chk) == 0)
return 0;
++chk;
}
return 1;
}
Once you have it working, then worry about performance (two of my favourite mantras are "Wrong is the least optimised state" and "Get it working first, then get it working fast").
The things you can look into for optimisations include, but are not limited to:
Eratosthenes sieve where, provided the range of primes isn't too large, it can greatly improve speed by not having to do a lot of calculations for each prime test; and
Using the fact that all primes other than two and three are of the form 6n±1, effectively tripling the speed of the isPrime function (see here for an explanation).
For that second bullet point, you can use:
int isPrime(unsigned int num) {
// Special cases for 0-3.
if (num < 2) return 0;
if (num < 4) return 1;
int chk = 5, add = 2; // prime generator, 6n +/- 1.
while (chk * chk <= num) // check every candidate.
if ((num % chk) == 0) // check if composite.
return 0;
chk += add; // next candidate.
add = 6 - add; // alternate +2, +4.
}
return 1; // no factors, must be prime.
}
Related
I am writing a program to see if a user entered number is Armstrong or not, here is my code:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int main(){
int x = 0;
printf("Enter a natural number: ");
scanf("%d", &x);
int ans = x;
// Digit Counter
int counter = 0; //Variable for number of digits in the user entered number
int b = x; //For each time number can be divided by 10 and isnt 0
for (int i = 1; i <= x; i++){ // Then counter variable is incremented by 1
b /= 10;
if (b != 0){
counter += 1;
}
}
++counter;
//Digit Counter
int sum = 0;
// Digit Finder
int D;
for (int j = 1; j <= x; j++){
D = x % 10; //Shows remainder of number (last digit) when divided by 10
sum += pow(D, counter); //Raises Digit found by counter and adds to sum
printf("%d\n", sum);
x /= 10; // Divides user entered number by 10 to get rid of digit found
}
if (sum == ans){
printf("%d is a Armstrong number! :)", ans);
}else
printf("%d is not an Armstrong number :(", ans);
//Digit Finder
return 0;
}
My problem is that the program works fine apart from one point, when the program is given a Armstrong number which does not start with 1 then it behaves normally and indicates if it is an Armstrong number or not, but when i input a Armstrong number which start with 1 then it will print out the Armstrong number but -1.
For example: If i input something such as 371 which is an Armstrong number it will show that it is an Armstrong number. However if i input 1634 it will output 1633 which is 1 less than 1634.
How can i fix this problem?, also by the way could someone comment on my code and tell me if it seems good and professional/efficient because i am a beginner in C and would like someone else's opinion on my code.
How can I fix this problem.
You know the number of iterations you want to make once you have calculated the digit count. So instead of looping till you reach the value of x:
for (int j = 1; j <= x; j++){
use the digit counter instead:
for (int j = 1; j <= counter; j++) {
also by the way could someone comment on my code and tell me if it seems good and professional/efficient because i am a beginner in C and would like someone else's opinion on my code.
There's a number of things you can do to improve your code.
First and foremost, any piece of code should be properly indented and formatted. Right now your code has no indenting, which makes it more difficult to read and it just looks ugly in general. So, always indent your code properly. Use an IDE or a good text editor, it will help you.
Be consistent in your code style. If you are writing
if (some_cond) {
...
}
else
//do this
It is not consistent. Wrap the else in braces as well.
Always check the return value of a function you use, especially for scanf. It will save you from many bugs in the future.
if (scanf("%d", &x) == 1)
//...all OK...
else
// ...EOF or conversion failure...
exit(EXIT_FAILURE);
Your first for loop will iterate x times uselessly. You can stop when you know that you have hit 0:
for (int i = 1; i <= x; i++){ // Then counter variable is incremented by 1
b /= 10;
if (b == 0){
break;
}
counter += 1;
}
C has ++ operator. Use that instead of doing counter += 1
int D; you create this, but don't initialize it. Always initialize your variables as soon as possible
C has const qualifier keyword, which makes a value immutable. This makes your code more readable, as the reader can immediately tell that this value will not change. In your code, you can change ans variable and make it a const int because it never changes:
const int ans = x;
Use more descriptive names for your variables. ans, D don't tell me anything. Use proper names, so that the reader of your code can easily understand your code.
These are some of the things that in my opinion you should do and keep doing to improve your code and coding skills. I am sure there can be more things though. Keep your code readable and as simple as possible.
The condition in this loop
for (int i = 1; i <= x; i++){ // Then counter variable is incremented by 1
b /= 10;
if (b != 0){
counter += 1;
}
}
does not make sense because there will be numerous redundant iterations of the loop.
For example if x is equal to 153 that is contains only 3 digits the loop will iterate exactly 153 times.
Also additional increment of the variable counter after the loop
++counter;
makes the code logically inconsistent.
Instead of the loop you could write at least the following way
int counter = 0;
int b = x;
do
{
++counter;
} while ( b /= 10 );
This loop iterates exactly the number of times equal to the number of digits in a given number.
In this loop
for (int j = 1; j <= x; j++){
D = x % 10; //Shows remainder of number (last digit) when divided by 10
sum += pow(D, counter); //Raises Digit found by counter and adds to sum
printf("%d\n", sum);
x /= 10; // Divides user entered number by 10 to get rid of digit found
}
it seems you did not take into account that the variable x is decreased inside the body of the loop
x /= 10; // Divides user entered number by 10 to get rid of digit found
So the loop can interrupt its iterations too early. In any case the condition of the loop again does not make great sense the same way as the condition of the first loop and only adds a bug.
The type of used variables that store a given number should be unsigned integer type. Otherwise the user can enter a negative number.
You could write a separate function that checks whether a given number is an Armstrong number.
Here you are.
#include <stdio.h>
int is_armstrong( unsigned int x )
{
const unsigned int Base = 10;
size_t n = 0;
unsigned int tmp = x;
do
{
++n;
} while ( tmp /= Base );
unsigned int sum = 0;
tmp = x;
do
{
unsigned int digit = tmp % Base;
unsigned int power = digit;
for ( size_t i = 1; i < n; i++ ) power *= digit;
sum += power;
} while ( ( tmp /= Base ) != 0 && !( x < sum ) );
return tmp == 0 && x == sum;
}
int main(void)
{
unsigned int a[] =
{
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 153, 370, 371, 407,
1634, 8208, 9474, 54748, 92727, 93084, 548834
};
const size_t N = sizeof( a ) / sizeof( *a );
for ( size_t i = 0; i < N; i++ )
{
printf( "%u is %san Armstrong number.\n", a[i], is_armstrong( a[i] ) ? "": "not " );
}
return 0;
}
The program output is
0 is an Armstrong number.
1 is an Armstrong number.
2 is an Armstrong number.
3 is an Armstrong number.
4 is an Armstrong number.
5 is an Armstrong number.
6 is an Armstrong number.
7 is an Armstrong number.
8 is an Armstrong number.
9 is an Armstrong number.
153 is an Armstrong number.
370 is an Armstrong number.
371 is an Armstrong number.
407 is an Armstrong number.
1634 is an Armstrong number.
8208 is an Armstrong number.
9474 is an Armstrong number.
54748 is an Armstrong number.
92727 is an Armstrong number.
93084 is an Armstrong number.
548834 is an Armstrong number.
Please remove j++ from 2nd loop for (int j = 1; j <= x; j++)
I tried this:
void armstrong(int x)
{
// count digits
int counter = 0, temp = x, sum = 0;
while(temp != 0)
{
temp = temp/10;
++counter; // Note: pre increment faster
}
// printf("count %d\n",counter);
temp = x;
while(temp != 0)
{
sum += pow(temp % 10, counter);
temp = temp/10;
}
// printf("sum %d\n",sum);
if(x == sum)
{
printf("Armstrong\n");
}
else
{
printf("No Armstrong\n");
}
}
int main(){
armstrong(371);
armstrong(1634);
return 0;
}
Let's take this and add the ability to handle multiple numeric bases while we're at it. Why? BECAUSE WE CAN!!!! :-)
#include <stdio.h>
#include <math.h>
double log_base(int b, double n)
{
return log(n) / log((double)b);
}
int is_armstrong_number(int b, /* base */
int n)
{
int num_digits = trunc(log_base(b, (double)n)) + 1;
int sum = 0;
int remainder = n;
while(remainder > 0)
{
sum = sum + pow(remainder % b, num_digits);
remainder = (int) (remainder / b);
}
return sum == n;
}
int main()
{
printf("All the following are valid Armstrong numbers\n");
printf(" 407 base 10 - result = %d\n", is_armstrong_number(10, 407));
printf(" 0xEA1 base 16 - result = %d\n", is_armstrong_number(16, 0xEA1));
printf(" 371 base 10 - result = %d\n", is_armstrong_number(10, 371));
printf(" 1634 base 10 - result = %d\n", is_armstrong_number(10, 1634));
printf(" 0463 base 8 - result = %d\n", is_armstrong_number(8, 0463));
printf("All the following are NOT valid Armstrong numbers\n");
printf(" 123 base 10 - result = %d\n", is_armstrong_number(10, 123));
printf(" 0x2446 base 16 - result = %d\n", is_armstrong_number(16, 0x2446));
printf(" 022222 base 8 - result = %d\n", is_armstrong_number(8, 022222));
}
At the start of is_armstrong_number we compute the number of digits directly instead of looping through the number. We then loop through the digits of n in base b, summing up the value of the digit raised to the number of digits in the number, for the given numeric base. Once the remainder hits zero we know there are no more digits to compute and we return a flag indicating if the given number is an Armstrong number in the given base.
I'm a first year student in a programming university and my first assignment is to find the sum of prime numbers between 3990000000 and 4010000000. The problem is everything I do, when I run the program it says the sum is 0 with a return value of 25. I've been trying to debug this code but with no luck, could someone help me?
My code is:
#include <stdio.h>
#define STARTNUMBER 3990000000
#define ENDNUMBER 4010000000
int main() {
unsigned int num;
int j, c, flag, sum = 0;
flag = 1;
c = 5;
j = 7;
for (num = STARTNUMBER; num <= ENDNUMBER; num++) {
if (num % 2 == 0) { /*if number mod 2 equals zero go to next number*/
flag = 0;
break;
}
if (num % 3 == 0) { /*if number mod 3 equals zero go to next number*/
flag = 0;
break;
} else
/*check if number is prime with the sequences 5+6+6...<=sqrt(number) and 7+6+6..<=sqrt(number)*/
while (c * c <= num && j * j <= num && flag == 1) {
if (num % c == 0 || num % j == 0) {
flag = 0;
break;
}
c += 6;
j += 6;
}
if (flag == 1)
sum++;
}
printf("There are %d prime numbers", sum);
}
You are asking for the sum of the prime number, even if your code is just printing how many they are. Assuming you've misunderstood the exercise, I try to show a possible problem of your original question, glimpsing a possible trick in your exercise since the interval is very close to 232.
Assuming also you are on a 64-bit environment, if there are at least two prime numbers in that inteval, the sum is going to be greater than INT_MAX (231 - 1). An int is not sufficient to store a value, and also unsigned int is not sufficient since UINT_MAX is 232 - 1.
Finally, assuming you've solved your problems with the break statements already described in the comments, try to store your sum variable into a unsigned long int, and replace the last part of the loop with
if (flag==1)
sum += num;
I need to find the biggest divisor of a positive integer and output it. Divisor should not be 1 or be equal to the integer itself. If it's a prime number the output should be "0". I have this code so far. However it doesn't work. It only works when I use "break" instead of "return 0" statement, but according to the task I should not use break :( How can I fix it? Thnx
#include <stdio.h>
int main() {
int input, maxDiv;
int div = 2;
scanf("%d", &input);
for ( ; div <= input/2; div += 1 ) {
if ( input % div == 0 ) {
maxDiv = input / div;
return 0;
} else {
maxDiv = 0;
}
}
printf("%d\n", maxDiv);
return 0;
}
You can rewrite it this way
int main(){
int input, maxDiv = 0;
int div = 2;
scanf("%d", &input);
for(; !maxDiv; div++)
if(!(input%div))
maxDiv = input/div;
printf("%d\n", ( maxDiv == 1 || input < 0 ? 0 : maxDiv ) );
return 0;
}
It is an infinite loop that will exit as soon as maxDiv != 0. The complexity is O(sqrt (n)) as there is always a divisor of n less than or equal to sqrt(n), so the code is bound to exit (even if input is negative).
I forgot, you have to handle the case where input is zero.
Maybe you can declare a flag?
#include <stdio.h>
int main() {
int input, maxDiv;
int div = 2;
char found = 0;
scanf("%d", &input);
for ( ; div <= input/2 && !found ; div += 1 ) {
if ( input % div == 0 ) {
maxDiv = input / div;
found = 1;
} else {
maxDiv = 0;
}
}
printf("%d\n", maxDiv);
return 0;
}
You can stop the loop when you reach sqrt(input)... it's not that difficult to find a perfectly good integer sqrt function.
There's not a lot of point dividing by all the even numbers after 2. In fact there's not a lot of point dividing by anything except the primes. It's not hard to find the primes up to sqrt(INT_MAX) (46340, for 32-bit integer)... there are tables of primes freely available if you don't want to run a quick sieve to generate same.
And the loop...
maxdiv = 0 ;
i = 0 ;
sq = isqrt(input) ;
while ((maxdiv == 0) && (prime[i] < sq))
{
if ((input % prime[i]) == 0)
maxdiv = input / prime[i] ;
i += 1 ;
} ;
assuming a suitable integer sqrt function and a table of primes... as discussed.
Since you are looking for the largest divisor, is there a reason you're not looping backward to 2? If there isn't, then there should be no need for a break statement or any special logic to exit the loop as you should keep looping until div is greater than input / 2, testing every value until you find the largest divisor.
maxDiv = -1;
for (div = input / 2;
div >= 2 && maxDiv == -1;
--div)
{
if (input % div == 0)
maxDiv = div;
}
maxDiv += (maxDiv == -1);
printf ("%d\n", maxDiv);
I added the extra condition of maxDiv being -1, which is like adding a conditional break statement. If it is still -1 by the end of the loop, then it becomes 0 because maxDiv += 1 is like writing maxDiv = -1 + 1, which is 0.
Without any jump statement such as break, this sort of test is what you must do.
Also, regarding your code, if I input 40, the if statement will be triggered when div is 2, and the program will end. If the return 0 is changed to a break, maxDiv will be 2, not 20. Looping backward will find 20 since 40/2=20, and 40%20==0.
Let us denote D to the max divisor of a given composite number N > 1.
Then, obviously, the number d = N / D is the min non-trivial divisor of N.
If d would not a primer number, then d would have a non-trivial divisor p < d.
By transitivity, this implies that p is a divisor of N, but this fact would contradict the fact that d is the min divisor of N, since p < d.
So, d must be a prime number.
In particular, it is enouth to search over those numbers which are less than sqrt(N), since, if p is a prime number greater than sqrt(N) which divies N, then N / p <= sqrt(N) (if not, *p * (N / p) > sqrt(N)sqrt(N) == N, wich is absurd).
This shows that it's enough to do the search the least divisor d of N just within the range of primer numbers from 2 to sqrt(N).
For efficiency, the value sqrt(N) must be computed just once before the loop.
Moreover, it is enough a rough approximation of sqrt(N), so we can write:
#include <math.h>
#include <stdio.h>
int main(void)
{
int N;
scanf("%d",&N);
// First, we discard the case in that N is trivial
// 1 is not prime, but indivisible.
// Change this return if your want.
if (N == 1)
return 0;
// Secondly, we discard the case in that N is even.
if (N % 2 == 0)
return N / 2;
// Now, the least prime divisor of N is odd.
// So, we increment the counter by 2 in the loop, by starting in 3.
float sqrtN = fsqrt(N); // square root of N in float precision.
for(d = 3; d <= sqrtN; d += 2)
if (N % d == 0)
return N/d;
// If the loop has reached its end normally,
// it means that N is prime.
return 0;
}
I think that the problem is not well stated, since I consider that a better flag to signalize that N is prime would be a returned value of 1.
There are more efficient algorithms to determine primality, but they are beyond the scope of the present question.
int prime(unsigned long long n){
unsigned val=1, divisor=7;
if(n==2 || n==3) return 1; //n=2, n=3 (special cases).
if(n<2 || !(n%2 && n%3)) return 0; //if(n<2 || n%2==0 || n%3==0) return 0;
for(; divisor<=n/divisor; val++, divisor=6*val+1) //all primes take the form 6*k(+ or -)1, k[1, n).
if(!(n%divisor && n%(divisor-2))) return 0; //if(n%divisor==0 || n%(divisor-2)==0) return 0;
return 1;
}
The code above is something a friend wrote up for getting a prime number. It seems to be using some sort of sieving, but I'm not sure how it exactly works. The code below is my less awesome version. I would use sqrt for my loop, but I saw him doing something else (probably sieving related) and so I didn't bother.
int prime( unsigned long long n ){
unsigned i=5;
if(n < 4 && n > 0)
return 1;
if(n<=0 || !(n%2 || n%3))
return 0;
for(;i<n; i+=2)
if(!(n%i)) return 0;
return 1;
}
My question is: what exactly is he doing?
Your friend's code is making use of the fact that for N > 3, all prime numbers take the form (6×M±1) for M = 1, 2, ... (so for M = 1, the prime candidates are N = 5 and N = 7, and both those are primes). Also, all prime pairs are like 5 and 7. This only checks 2 out of every 3 odd numbers, whereas your solution checks 3 out of 3 odd numbers.
Your friend's code is using division to achieve something akin to the square root. That is, the condition divisor <= n / divisor is more or less equivalent to, but slower and safer from overflow than, divisor * divisor <= n. It might be better to use unsigned long long max = sqrt(n); outside the loop. This reduces the amount of checking considerably compared with your proposed solution which searches through many more possible values. The square root check relies on the fact that if N is composite, then for a given pair of factors F and G (such that F×G = N), one of them will be less than or equal to the square root of N and the other will be greater than or equal to the square root of N.
As Michael Burr points out, the friend's prime function identifies 25 (5×5) and 35 (5×7) as prime, and generates 177 numbers under 1000 as prime whereas, I believe, there are just 168 primes in that range. Other misidentified composites are 121 (11×11), 143 (13×11), 289 (17×17), 323 (17×19), 841 (29×29), 899 (29×31).
Test code:
#include <stdio.h>
int main(void)
{
unsigned long long c;
if (prime(2ULL))
printf("2\n");
if (prime(3ULL))
printf("3\n");
for (c = 5; c < 1000; c += 2)
if (prime(c))
printf("%llu\n", c);
return 0;
}
Fixed code.
The trouble with the original code is that it stops checking too soon because divisor is set to the larger, rather than the smaller, of the two numbers to be checked.
static int prime(unsigned long long n)
{
unsigned long long val = 1;
unsigned long long divisor = 5;
if (n == 2 || n == 3)
return 1;
if (n < 2 || n%2 == 0 || n%3 == 0)
return 0;
for ( ; divisor<=n/divisor; val++, divisor=6*val-1)
{
if (n%divisor == 0 || n%(divisor+2) == 0)
return 0;
}
return 1;
}
Note that the revision is simpler to understand because it doesn't need to explain the shorthand negated conditions in tail comments. Note also the +2 instead of -2 in the body of the loop.
He's checking for the basis 6k+1/6k-1 as all primes can be expressed in that form (and all integers can be expressed in the form of 6k+n where -1 <= n <= 4). So yes it is a form of sieving.. but not in the strict sense.
For more:
http://en.wikipedia.org/wiki/Primality_test
In case the 6k+-1 portion is confusing, note that you can perform some factorization of most forms of 6k+n and some are obviously composite and some need to be tested.
Consider numbers:
6k + 0 -> composite
6k + 1 -> not obviously composite
6k + 2 -> 2(3k+1) --> composite
6k + 3 -> 3(2k+1) --> composite
6k + 4 -> 2(3k+2) --> composite
6k + 5 -> not obviously composite
I've not seen this little trick before, so it's neat, but of limited utility since a sieve of Eratosthenese is more efficient for finding many small prime numbers, and larger prime numbers benefit from faster, more intelligent, tests.
#include<stdio.h>
int main()
{
int i,j;
printf("enter the value :");
scanf("%d",&i);
for (j=2;j<i;j++)
{
if (i%2==0 || i%j==0)
{
printf("%d is not a prime number",i);
return 0;
}
else
{
if (j==i-1)
{
printf("%d is a prime number",i);
}
else
{
continue;
}
}
}
}
#include<stdio.h>
int main()
{
int n, i = 3, count, c;
printf("Enter the number of prime numbers required\n");
scanf("%d",&n);
if ( n >= 1 )
{
printf("First %d prime numbers are :\n",n);
printf("2\n");
}
for ( count = 2 ; count <= n ; )
{
for ( c = 2 ; c <= i - 1 ; c++ )
{
if ( i%c == 0 )
break;
}
if ( c == i )
{
printf("%d\n",i);
count++;
}
i++;
}
return 0;
}
I am trying to generate all the prime factors of a number n. When I give it the number 126 it gives me 2, 3 and 7 but when I give it say 8 it gives me 2, 4 and 8. Any ideas as to what I am doing wrong?
int findPrime(unsigned long n)
{
int testDivisor, i;
i = 0;
testDivisor = 2;
while (testDivisor < n + 1)
{
if ((testDivisor * testDivisor) > n)
{
//If the test divisor squared is greater than the current n, then
//the current n is either 1 or prime. Save it if prime and return
}
if (((n % testDivisor) == 0))
{
prime[i] = testDivisor;
if (DEBUG == 1) printf("prime[%d] = %d\n", i, prime[i]);
i++;
n = n / testDivisor;
}
testDivisor++;
}
return i;
}
You are incrementing testDivisor even when you were able to divide n by it. Only increase it when it is not divisible anymore. This will result in 2,2,2, so you have to modify it a bit further so you do not store duplicates, but since this is a homework assignment I think you should figure that one out yourself :)
Is this based on an algorithm your professor told you to implement or is it your own heuristic? In case it helps, some known algorithms for prime factorization are the Quadratic Sieve and the General Number Field Sieve.
Right now, you aren't checking if any divisors you find are prime. As long as n % testDivisor == 0 you are counting testDivisor as a prime factor. Also, you are only dividing through by testDivisor once. You could fix this a number of ways, one of which would be to replace the statement if (((n % testDivisor) == 0)) with while (((n % testDivisor) == 0)).
Fixing this by adding the while loop also ensures that you won't get composite numbers as divisors, as if they still divide n, a smaller prime factor must have also divided n and the while loop for that prime factor wouldn't have left early.
Here is code to find the Prime Factor:
long GetPrimeFactors(long num, long *arrResult)
{
long count = 0;
long arr[MAX_SIZE];
long i = 0;
long idx = 0;
for(i = 2; i <= num; i++)
{
if(IsPrimeNumber(i) == true)
arr[count++] = i;
}
while(1)
{
if(IsPrimeNumber(num) == true)
{
arrResult[idx++] = num;
break;
}
for(i = count - 1; i >= 0; i--)
{
if( (num % arr[i]) == 0)
{
arrResult[idx++] = arr[i];
num = num / arr[i];
break;
}
}
}
return idx;
}
Reference: http://www.softwareandfinance.com/Turbo_C/Prime_Factor.html
You can use the quadratic sieve algorithm, which factors 170-bit integers in second and 220-bit integers in minute. There is a pure C implementation here that does not require GMP or an external library : https://github.com/michel-leonard/C-Quadratic-Sieve, it's able to provide you a list of the prime factors of N. Thank You.