Розбір

Let's construct the required sequence in a constructive manner. Since the sine function is periodic with a period of $2 π$ , we will look for it in the form of $s in (k (2 π + ɛ))$ , where $k = 0, 1, 2, ...$ . We need to find such a positive integer $A$ for which there exists an integer $k$ such that $A = 2 πk + ɛ$ , where $ɛ$ is as small as possible. For example, if we check values of $A$ from $1$ to $1000$ , the smallest $ε$ is achieved when $A = 710$ . In this case, $710 \approx 2 π \cdot 113 + 0.00006028$ .

Let $A = 710$ . Consider the sequence

s in (A), s in (2 A), s in (3 A), ..., s in (n A)

The values of the elements in this sequence are

s in (2 π \cdot 113 + ɛ), s in (2 \cdot 2 π \cdot 113 + 2 \cdot ɛ), s in (3 \cdot 2 π \cdot 113 + 3 \cdot ɛ), ..., s in (n \cdot 2 π \cdot 113 + n \cdot ɛ)

or, considering the periodicity, these values are equivalent to

s in (ɛ), s in (2 \cdot ɛ), s in (3 \cdot ɛ), \dots, s in (n \cdot ɛ)

The function $s in (x)$ is increasing on the interval $x \in [0; π /2]$ . For the sequence of the above real numbers to be monotonically increasing, it is sufficient to satisfy the inequality: $n \cdot ɛ \leq π /2$ . Substituting $n = 10000$ and $ɛ = 0.00006028$ , we get: $10000 \cdot 0.00006028 \leq π /2$ or $0.6028 \leq 1.57$ , which is true. Therefore, by this method, it is possible to find $1 0^{4}$ required integers.

Algorithm realization

Read the input value of $n$ .

scanf("%d", &n);

On the interval $[1..1000]$ , we find such a positive integer $A$ for which, when represented as $A = 2 πk + ɛ$ , the value of $ε$ will be the smallest.

bestA = 0;
bestE = 2 * PI;
for (a = 1; a < 1000; a++)
{
  e = fmod(a, 2.0 * PI);
  if (e < bestE)
  {
    bestE = e;
    bestA = a;
  }
}

The value of $x = b es t A$ will be $710$ .

x = bestA;

Print the desired sequence: $0 \cdot 710, 1 \cdot 710, 2 \cdot 710, ..., (n - 1) \cdot 710$ .

for (i = 0; i < n; i++)
  printf("%d ", x * i);
printf("\n");