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  1. #include <stdio.h>
  2.  
  3. #define PRIMES_COUNT 37
  4. #define MAX_PRIMES 10
  5.  
  6. int main() {
  7. 	static int primes[PRIMES_COUNT] = {
  8.         2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53,
  9.         59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113,
  10.         127, 131, 137, 139, 149, 151, 157
  11.     };
  12.  
  13. 	unsigned int k;
  14.     int i, n;
  15.     int current_prime = 0, answer = -1;
  16.     unsigned int temp;
  17.     int key_index;
  18.  
  19.     int multK[MAX_PRIMES];
  20.     int keys[MAX_PRIMES];
  21.     int multN[MAX_PRIMES];
  22.  
  23. 	for (i = 0; i < MAX_PRIMES; i++) {
  24.         multK[i] = 0;
  25.         keys[i]=0;
  26.         multN[i] = 0;
  27. 	}
  28.  
  29.     scanf("%d%u", &n, &k);
  30.  
  31. 	temp = k;
  32. 	key_index=0;
  33.  
  34. 	int increase_key = 0;
  35.  
  36. 	while (current_prime < PRIMES_COUNT)
  37.         if (temp % primes[current_prime] == 0) {
  38.             temp /= primes[current_prime];
  39.             if ( multK[key_index]==0 )
  40.             {
  41.             	keys[key_index]=primes[current_prime];
  42.             	increase_key = 1;
  43.             }
  44.             multK[key_index]++;
  45.         } else 
  46.         {
  47.         	current_prime++;
  48.  
  49.         	if (increase_key==1)
  50.         	{
  51.         		increase_key = 0;
  52.         		key_index++;
  53.         	}
  54.  
  55.         	if (key_index>=MAX_PRIMES)
  56.         	{
  57.         		printf("Слишком маленький размер массива ключей");
  58.         		exit(1);
  59.         	}
  60.         }
  61.  
  62. 	for (i = 2; i <= n; i++) {
  63.         int temp = i;
  64. 		key_index=0;
  65. 		while (key_index<MAX_PRIMES && multK[key_index]!=0)
  66. 			if (temp % keys[key_index] == 0) {
  67.                     temp /= keys[key_index];
  68.                     multN[key_index]++;
  69.                 } else key_index++;
  70.  
  71.     }
  72.  
  73.     for (i = 0; i < MAX_PRIMES && multK[i]!=0; i++) {
  74.         temp = multN[i] / multK[i];
  75.         if (temp < answer || answer == -1) answer = temp;
  76.     }
  77.  
  78.     printf ("%i\n", answer);
  79.     return (0);
  80. }
Success #stdin #stdout 0s 2296KB
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https://ideone.com/XbEHZ2
language:
C (gcc 8.3)
created:
11 years ago
visibility:
public

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