Need to check whether the number is simple using the function.
I decided that way, but it works not to the end correctly. I can not catch the reason …

int prime (unsigned int num {
  For (int i = 2; I * i & lt; = num; i ++) {
    if (num% i == 0) {
      Return 0;
    }
  }
  Return 1;
}

Answer 1, Authority 100%

bool is_prime (unsigned p) {
  If (P & LT; 2) Return False;
  if (p == 2) Return True;
  if (p% 2 == 0) Return False;
  Double Limit = SQRT (P);
  For (unsigned i = 3; i & lt; = limit; i + = 2) {
    if ((p% i) == 0) Return False;
  }
  RETURN TRUE;
}

Answer 2

Small optimization. First, check the num equal to 2 (simple), less than 2 and num even number (not simple). Then you can only sore odd:

uint32_t i, lim = sqrt (num) + 1;
for (i = 3; i & lt; lim; i + = 2)
 if (num% i == 0)
  Return 0;
Return 1;

looked asm. code. Despite -o3 a little better:

uint32_t i, lim = ((uint32_t) SQRT (NUM)) + 1;

–

@vladd , it’s not for this task. The sieve is the search all simple (specified them the number of ). Here you need to check only one number.

Answer 3

Here’s the solution on C++. You want to rewrite onto a Source: Akina Reshelo

# include & lt; iostream & gt;
#Include & lt; vector & gt;
Using Namespace STD;
BOOL IS_PRIME (unsigned int num) {
int limit = num;
INT SQR_LIM;
Vector & lt; bool & gt; IS_PRIME_F (Num + 10);
int x2, y2;
int i, j;
int n;
// Initialization of the sieve
SQR_LIM = (INT) SQRT ((Long Double) Limit);
for (i = 0; i & lt; = limit; i ++) is_prime_f [i] = false;
IS_PRIME_F [2] = TRUE;
is_prime_f [3] = True;
// Presumably simple - it is integer with an odd number
// Views in these square forms.
// X2 and Y2 are squares I and J (optimization).
x2 = 0;
For (i = 1; i & lt; = SQR_LIM; I ++) {
  x2 + = 2 * i - 1;
  y2 = 0;
  For (j = 1; j & lt; = sqr_lim; j ++) {
    y2 + = 2 * j - 1;
    n = 4 * x2 + y2;
    if ((N & LT; = Limit) & amp; & amp; (n% 12 == 1 || n% 12 == 5))
      is_prime_f [n] =! is_prime_f [n];
    // n = 3 * x2 + y2;
    n - = x2; // Optimization
    if ((N & LT; = Limit) & amp; & amp; (n% 12 == 7))
      is_prime_f [n] =! is_prime_f [n];
    // n = 3 * x2 - y2;
    n - = 2 * y2; // Optimization
    if ((i & gt; j) & amp; & amp; (n & lt; = limit) & amp; & amp; (n% 12 == 11))
      is_prime_f [n] =! is_prime_f [n];
  }
}
// Count the multiple squares of prime numbers in the interval [5, SQRT (Limit)].
// (the main stage can not cut them out)
For (i = 5; I & lt; = SQR_LIM; I ++) {
  if (IS_PRime_F [i]) {
    n = i * i;
    For (j = n; j & lt; = limit; j + = n) {
      IS_PRIME_F [J] = false;
    }
  }
}
// Conclusion of a list of prime numbers into the console.
// Printf ("2, 3, 5");
if ((i == 2) || (i == 3) || (i == 5))
  RETURN TRUE;
For (i = 6; i & lt; = limit; I ++) {// Added a varying test at 3 and 5. In the original version of the algorithm, there is no need for it.
  if ((IS_PRIME_F [i]) & amp; & amp; (i% 3! = 0) & amp; & amp; (i% 5! = 0)) {
    // Printf (",% d", i);
       if (i == num)
         RETURN TRUE;
  }
}
  Return 0;
}
Int Main (int argc, char * argv []) {
  if (IS_PRime (983)) {
    COUT & LT; & LT; "Prime" & lt; & lt; Endl;
  } else {
    COUT & LT; & LT; "NOT PRIME" & LT; & LT; Endl;
  }
  Return 0;
}

Answer 4

I agree with the Renegator. For VBA (perhaps someone needs) String function, Returning all squares of prime numbers