I have to make a program that displays all the proper divisors of a number(given by the user),basically all except the number itself and 1. If it doesn't have any print that it's prime.All good,done that.
There is one more thing I need to do though in case it does have proper divisors and that is to display the smallest and biggest divisor which I can't figure out how to do ,I tried to do it as if the printed divisors were just one number ignoring the "\n" and using "number % 10" to find the last number and the while loop to find the first number,but that won't work in case let's say the given number is 33 and the biggest divisor is 11.
I'll provide the code for better understanding since my question is kinda fuzzy. Everything is how I want it to be except I don't know how to display the smallest divisor and the biggest divisor.
#include <stdio.h>
int main() {
int n, divisor, smallest, biggest, count = 0;
scanf("%d", &n);
for (divisor = 2; divisor < n; divisor++) {
if (n % divisor == 0) {
printf("%d is divisor of %d\n", divisor, n);
}
}
for (divisor = 1; divisor <= n; divisor++) {
if (n % divisor == 0) {
count++;
}
}
if (count == 2) {
printf("The number does not have proper divisors(it is prime)");
}
return 0;
}
I would modify your code as follows:
#include <stdio.h>
int main()
{
int n,divisor,smallest,biggest,count=0;
scanf("%d",&n);
for(divisor = 2;divisor < n;divisor++)
{
if(n % divisor == 0)
{
count++;
if (count == 1) smallest = divisor;
biggest = divisor;
printf("%d is divisor of %d\n",divisor,n);
}
}
if (count == 0) printf("The number does not have proper divisors(it is prime)");
else printf("%d and %d are the smallest and biggest divisors respectively", smallest, biggest);
return 0;
}
Instead of using two for loops, I merged them into one since they're doing the same thing. Every positive number will be divisible by 1 and itself so it's redundant to use the second for loop. When count is 1, I know that I have found the first divisor so I set smallest equal to that. Also,everytime I find a divisor, I set biggest equal to that so that in the end, biggest is set to the last divisor found (note that you should not use an else clause here, since that will cause the program to not work when the smallest and biggest divisors are equal). I also increment count by 1 everytime a divisor is found in order to check whether the number is prime without modifying your approach too much (there are efficient ways to check for prime numbers, and I encourage you to take a look at them). And finally, I print the smallest and biggest divisors only if they exist, else I print that it is prime.
You can implement minDivisor only, then to find max divisor you can divide value by min divisor:
#include <math.h>
...
/* Returns the smallest non-trivial divisor or 1 if it doesn't exist */
int minDivisor(int value) {
if (value <= 3)
return 1;
/* let's check n for being even */
if (value % 2 == 0)
return 2;
/* We should check up to sqrt(n) only */
int n = (int) (sqrt(value) + 0.5);
for (int divisor = 3; divisor <= n; divisor += 2)
if (value % divisor == 0)
return divisor;
/* prime number */
return 1;
}
Then
int main() {
int n, divisor, smallest, biggest, count = 0;
...
smallest = minDivisor(n);
biggest = n / smallest;
...
Please, fiddle yourself
Related
I'm trying to create a program that breaks down a number into its component digits.
The code I've written so far is as follows:
#include <stdio.h>
int main() {
int num;
int digits;
int count = 1;
printf("Insert a positive number: ");
do {
scanf("%d", &num);
} while (num <= 0);
printf("In digits... \n");
// Number in digits
while (num != 0) {
digits = num % 10;
printf("Digit %d --> %d \n", count, digits);
num /= 10;
count++;
}
}
The digits of the number are printed correctly, but in reverse order!
Insert a positive number: 345
In digits...
Digit 1 --> 5
Digit 2 --> 4
Digit 3 --> 3
I can't figure out how to fix this, can anyone help me?
You're printing the number mod 10, which is the last digit, then dividing the number by 10 and repeating until the number is zero. So it prints the digits from right to left.
If you want to print from left to right, you need to print the digit that has the highest power of ten first. Here's a naive way to do that, by first finding the highest power of ten the number has a digit for and then using a for loop to go from that power to one to print the digits from left to right:
void print_digits(int n) {
int mask = 1;
for(int n2 = n; n2; n2 /= 10) mask *= 10; // find the left-most power of ten
for(int i = 1; mask > 1; mask /= 10) // loop over the mask to 1
printf("Digit %d --> %d\n", i++, (n % mask) * 10 / mask);
// print the digit number and increment the digit counter
// extract and print the digit:
// `n % mask` gets rid of everything to the left
// `* 10 / mask` gets rid of everything to the right
}
You could also make a simpler solution using the standard library function sprintf (string print formatted) to put the int into a string and then print from that, like so:
void print_digits(int n) {
char num[11]; // 32-bit int up to 9 digits, possible '-', and \0 -> 11
sprintf(num, "%d", n);
for (int i = 0; num[i]; i++)
printf("Digit %d --> %c\n", i + 1, num[i]);
}
The second might also be a tiny bit more performant due to not involving division, but I'm not certain of that and such minor differences don't matter for a problem like this anyway
Your code prints them in reverse because you are starting with the right-most digit (by taking the modulo by 10) each time, then dividing by 10.
To print from left to right, you could have:
#include <stdint.h>
void print_uint32_digits(uint32_t val)
{
int started = 0; // Flag to say we have started printing digits
// Special case for when val is zero
if (val == 0u)
{
printf("0\n");
return;
}
for (uint32_t divider = 1000000000u; divider != 0u; divider /= 10u)
{
uint32_t num = val / divider;
if (num > 0u || started)
{
printf("%c", num + '0');
started = 1;
val -= (num * divider);
}
}
printf("\n");
}
Some notes - I've used uint32_t because it is unsigned and a nice length. To make it work with plain "ints" would involve handling the case of negative numbers, which is more complex. I don't know if you need to handle negative numbers. (Also, ints on your platform could be 64-bit, so the initial divider would have to be adjusted accordingly.)
The 'started' flag is used to signal when we have output at least one digit, to ensure zeros get printed correctly. Until that flag is set, leading zeros are not printed.
A very simple recursive solution looks like this:
void print_num(int num)
{
if (num < 10)
fputc('0' + num, stdout);
else {
print_num(num/10);
print_num(num%10);
}
}
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.
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.
I am accepting a composite number as an input. I want to print all its factors and also the largest prime factor of that number. I have written the following code. It is working perfectly ok till the number 51. But if any number greater than 51 is inputted, wrong output is shown. how can I correct my code?
#include<stdio.h>
void main()
{
int i, j, b=2, c;
printf("\nEnter a composite number: ");
scanf("%d", &c);
printf("Factors: ");
for(i=1; i<=c/2; i++)
{
if(c%i==0)
{
printf("%d ", i);
for(j=1; j<=i; j++)
{
if(i%j > 0)
{
b = i;
}
if(b%3==0)
b = 3;
else if(b%2==0)
b = 2;
else if(b%5==0)
b = 5;
}
}
}
printf("%d\nLargest prime factor: %d\n", c, b);
}
This is a bit of a spoiler, so if you want to solve this yourself, don't read this yet :). I'll try to provide hints in order of succession, so you can read each hint in order, and if you need more hints, move to the next hint, etc.
Hint #1:
If divisor is a divisor of n, then n / divisor is also a divisor of n. For example, 100 / 2 = 50 with remainder 0, so 2 is a divisor of 100. But this also means that 50 is a divisor of 100.
Hint #2
Given Hint #1, what this means is that we can loop from i = 2 to i*i <= n when checking for prime factors. For example, if we are checking the number 100, then we only have to loop to 10 (10*10 is <= 100) because by using hint #1, we will get all the factors. That is:
100 / 2 = 50, so 2 and 50 are factors
100 / 5 = 20, so 5 and 20 are factors
100 / 10 = 10, so 10 is a factor
Hint #3
Since we only care about prime factors for n, it's sufficient to just find the first factor of n, call it divisor, and then we can recursively find the other factors for n / divisor. We can use a sieve approach and mark off the factors as we find them.
Hint #4
Sample solution in C:
bool factors[100000];
void getprimefactors(int n) {
// 0 and 1 are not prime
if (n == 0 || n == 1) return;
// find smallest number >= 2 that is a divisor of n (it will be a prime number)
int divisor = 0;
for(int i = 2; i*i <= n; ++i) {
if (n % i == 0) {
divisor = i;
break;
}
}
if (divisor == 0) {
// we didn't find a divisor, so n is prime
factors[n] = true;
return;
}
// we found a divisor
factors[divisor] = true;
getprimefactors(n / divisor);
}
int main() {
memset(factors,false,sizeof factors);
int f = 1234;
getprimefactors(f);
int largest;
printf("prime factors for %d:\n",f);
for(int i = 2; i <= f/2; ++i) {
if (factors[i]) {
printf("%d\n",i);
largest = i;
}
}
printf("largest prime factor is %d\n",largest);
return 0;
}
Output:
---------- Capture Output ----------
> "c:\windows\system32\cmd.exe" /c c:\temp\temp.exe
prime factors for 1234:
2
617
largest prime factor is 617
> Terminated with exit code 0.
I presume you're doing this to learn, so I hope you don't mind a hint.
I'd start by stepping through your algorithm on a number that fails. Does this show you where the error is?
You need to recode so that your code finds all the prime numbers of a given number, instead of just calculating for the prime numbers 2,3, and 5. In other words, your code can only work with the number you are calculating is a prime number or is a multiple of 2, 3, or 5. But 7, 11, 13, 17, 19 are also prime numbers--so your code should simply work by finding all factors of a particular number and return the largest factor that is not further divisible.
Really, this is very slow for all but the smallest numbers (below, say, 100,000). Try finding just the prime factors of the number:
#include <cmath>
void addfactor(int n) {
printf ("%d\n", n);
}
int main()
{
int d;
int s;
int c = 1234567;
while (!(c&1)) {
addfactor(2);
c >>= 1;
}
while (c%3 == 0) {
addfactor(3);
c /= 3;
}
s = (int)sqrt(c + 0.5);
for (d = 5; d <= s;) {
while (c % d == 0) {
addfactor(d);
c /= d;
s = (int)sqrt(c + 0.5);
}
d += 2;
while (c % d == 0) {
addfactor(d);
c /= d;
s = (int)sqrt(c + 0.5);
}
d += 4;
}
if (c > 1)
addfactor(c);
return 0;
}
where addfactor is some kind of macro that adds the factor to a list of prime factors. Once you have these, you can construct a list of all the factors of the number.
This is dramatically faster than the other code snippets posted here. For a random input like 10597959011, my code would take something like 2000 bit operations plus 1000 more to re-constitute the divisors, while the others would take billions of operations. It's the difference between 'instant' and a minute in that case.
Simplification to dcp's answer(in an iterative way):
#include <stdio.h>
void factorize_and_print(unsigned long number) {
unsigned long factor;
for(factor = 2; number > 1; factor++) {
while(number % factor == 0) {
number = number / factor;
printf("%lu\n",factor);
}
}
}
/* example main */
int main(int argc,char** argv) {
if(argc >= 2) {
long number = atol(argv[1]);
factorize_and_print(number);
} else {
printf("Usage: %s <number>%<number> is unsigned long", argv[0]);
}
}
Note: There is a number parsing bug here that is not getting the number in argv correctly.