Sieve of Eratosthenes and his primes - c

Question

This is my code:
#include <stdio.h>
int main() {
int number;
int prime[200000] = { 0 };
int i = 0;
int j = 0;
int number1[200] = { 0 };
int t = 0;
int count = 0;
int newprime2[200][200];
int counter[200] = { 0 };
int square;
int count1;
while ((scanf("%d", &number) == 1 ) && (number != 0)) {
number1[count] = number;
++count;
}
count1 = count;
for (count = 0; count < count1; ++count) {
if (number1[count] < 0) {
fprintf(stderr, "Error: Invalid input!\n");
return 100;
break;
}
for (i = 0; i < number1[count]; i++) {
prime[i] = i;
}
for (i = 2; (i < (number1[count])); i++) {
if (prime[i] != 0) {
for (j = 2; (j < (number1[count])); j++) {
{
prime[j*prime[i]] = 0;
if (prime[i] * j > (number1[count]))
break;
}
}
}
}
t = 0;
for (i = 2; i < number1[count]; ++i) {
if ((prime[i] != 0) && (number1[count] % prime[i] == 0)) {
newprime2[count][t] = prime[i];
++t;
}
}
printf("\n");
printf("%i is made out of these primes\n", number1[count]);
counter[count] = 0;
square = 0;
for (i = 0; i < t; ++i) {
while (number1[count] % newprime2[count][i] == 0) {
number1[count] = number1[count] / newprime2[count][i];
square++;
}
counter[count]++;
/* if number isn't made out of any of these primes*/
if (!newprime2[count][i]) { /*Why is this not working?*/
printf("%i ", number1[count]);
}
if (counter[count] == 1) {
printf("%i^%d ", newprime2[count][i], square);
} else {
printf("* %i^%d ", newprime2[count][i], square);
}
square = 0;
}
}
printf("\n");
return 0;
}
For example, my input is: 1 11 120 8 0
Output looks like this:
1 is made out of these primes
11 is made out of these primes
120 is made out of these primes
2^3 * 3^1 * 5^1
8 is made out of these primes
2^3
But Output should looks like this:
1 is made out of these primes
1
11 is made out of these primes
11
...
Statement (!newprime2[count][i]) means that this array is empty right? So why it isn't working? And why I even can't use gcc -pedantic -Wall -Werror -std=c99 -O3 ? Can someone help me?

Answer 1

See this part of your code:
t = 0;
for (i = 2; i < number1[count]; ++i){
if ((prime[i]!=0) && (number1[count] % prime[i]==0)){
newprime2[count][t] = prime[i];
++t;
}
If number1[count] is 1, then the body of the for loops will not execute, sot will keep its value (0). Consequently the body of the next loop
for (i=0; i < t; ++i){
will not execute, too.
For number 11 the body of this loop will execute but it will do nothing as the condition in the if statement will be always false. So it results to the same problem - t will keep its value 0 with the same consequence.

Answer 2

The line
if (!newprime2[count][i])
is not reached if t==0 before the for-loop and that is the case if the input is prime or unity. Just check t and end there if it is zero.
Or check earlier if it is unity or it is in prime already.
I cannot repeat your problems with gcc -pedantic -Wall -Werror -std=c99 -O3.

Answer 3

Your algorithm is both too complicated and approximate:
You do not need to perform a sieve to factorize the numbers, you can just enumerate divisors, composite divisors will have a non zero remainder because their prime factors will have been removed already.
The sieve is incomplete: you go to 200000 which would be overkill if int type is 32 bits (46341 would suffice) and would be too small if int is 64 bits.
Here is a simplified version:
#include <stdio.h>
int main(void) {
int number, i, p, n, factors, count;
int numbers[200];
for (count = 0; count < 200 && scanf("%d", &number) == 1; count++) {
if (number == 0)
break;
if (number < 0) {
fprintf(stderr, "Error: Invalid input!\n");
return 100;
}
numbers[count] = number;
}
for (i = 0; i < count; i++) {
number = numbers[i];
printf("%d is made out of these primes\n", number);
factors = 0;
for (p = 2; p * p <= number; p += 1 + (p & 1)) {
if (number % p == 0) {
n = 0;
factors++;
do {
number /= p;
n++;
} while (number % p == 0);
if (n == 1)
printf("%d ", p);
else
printf("%d^%d ", p, n);
}
}
if (factors == 0 || number != 1)
printf("%d", number);
printf("\n");
}
return 0;
}