#include "../next_prime.h" 

uint8_t get_mask(uint32_t n) {
  switch(n % 30) {
    case 1:  return 1;
    case 7:  return 2;
    case 11: return 4;
    case 13: return 8;
    case 17: return 16;
    case 19: return 32;
    case 23: return 64;
    case 29: return 128;
    default: return 0; // делится на 2,3 или 5...
  }
}

#define TEST 0xFFFFFFFF

int main(int argc, char* argv[])
{
	int size = TEST / 30 + 1;
	uint8_t* bitmap = (uint8_t*)calloc(size, 1);
	uint32_t x = 7;
	while(x && x < TEST) {
		bitmap[x/30] |= get_mask(x);
		x = next_prime(x);
	}
	FILE* f = fopen("bitmap30.bin", "wb");
	fwrite(bitmap, size, 1, f);
	fclose(f);

	printf("bitmap ready size = %d\n", size);
	x = 7;
	for(uint32_t i = 0; i < TEST; i++) {
		if((bitmap[i/30] & get_mask(i))) {
			if(i != x) {printf("error %u != %u\n", i, x);break;}
			x = next_prime(x);
		}
	}
	printf("check end\n");

	any_key();
}

...
		// для определения номера текущего шага
		static uint32_t offset[] = {9,0,9,9,9,9,9,1,9,9,9,2,9,3,9,9,9,4,9,5,9,9,9,6,9,9,9,9,9,7};
		// порядок шагов
		static uint32_t step5[] = {6,4,2,4,2,4,6,2};
		uint32_t step_id = offset[max_prime % 30]; // max_prime последнее найденное простое
		for(uint32_t i = max_prime + step5[step_id]; i < limit; i += step5[step_id]) {
			step_id++;
			step_id &= 7; //if(step_id == 8) step_id = 0;
			cnt++;
			cnt2++;
...
		}
	}
	free(bitmap);
	printf("eratosfen5() %dM(%dM) steps\n", (uint32_t)(cnt / 1000000), (uint32_t)(cnt2 / 1000000));