我在学校有一项任务，在2分钟内计算最多10 ^ 18的素数，并且使用不超过2 GB的内存。我第一次尝试使用以下优化实现了分段筛：
问题是，为了计算我的计算机上的素数高达10 ^ 9（具有相当不错的规格）需要13秒，因此10 ^ 18会花费我几天。
我的问题是，是否有一些我缺少的优化，或者是否有更好更快的方法将素数计算到数字？代码：
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>
#include <string.h>
typedef signed char int8_t;
typedef signed short int int16_t;
typedef signed int int32_t;
typedef signed long long int int64_t;
typedef unsigned char uint8_t;
typedef unsigned short int uint16_t;
typedef unsigned int uint32_t;
typedef unsigned long long int uint64_t;
#define SIZE 32
#define DEBUG
#define KRED "\x1B[31m"
#define KGRN "\x1B[32m"
#define KYEL "\x1B[33m"
#define KBLU "\x1B[34m"
#define KMAG "\x1B[35m"
#define KCYN "\x1B[36m"
#define KWHT "\x1B[37m"
#define RESET "\033[0m"
struct node {
uint64_t data;
struct node* next;
};
struct queue {
struct node* first;
struct node* last;
uint32_t size;
};
typedef struct node Node;
typedef struct queue Queue;
/* Queue model */
uint8_t enqueue(Queue* queue, int64_t value) {
Node* node = (Node*)malloc(sizeof(Node*));
if (node == NULL)
return 0;
node->data = value;
if (queue->last)
queue->last->next = node;
queue->last = node;
if (queue->first == NULL)
queue->first = queue->last;
queue->size++;
return 1;
}
uint64_t dequeue(Queue* queue) {
Node* node = queue->first;
uint64_t save_data = node->data;
if (queue->size == 0)
return 0;
queue->first = queue->first->next;
queue->size--;
free(node);
return save_data;
}
Node* queue_peek(Queue* queue) {
return queue->first;
}
uint32_t queue_size(Queue* queue) {
return queue->size;
}
Queue* init_queue() {
Queue* queue = (Queue*)malloc(sizeof(Queue*));
queue->first = queue->last = NULL;
queue->size = 0;
return queue;
}
/* Working with bit arrays functions */
uint8_t count_set_bits(uint64_t nbr) {
uint8_t count = 0;
while (nbr) {
count++;
nbr &= (nbr - 1);
}
return count;
}
uint8_t get_bit(uint32_t array[], uint32_t position) {
const uint64_t mask = 1U << (position % SIZE);
return array[position / SIZE] & mask ? 1 : 0;
}
void clear_bit(uint32_t array[], uint32_t position) {
const uint64_t mask = ~(1U << (position % SIZE));
array[position / SIZE] &= mask;
}
void set_bit(uint32_t array[], uint32_t position) {
array[position / SIZE] |= (1U << (position % SIZE));
}
/* Solve the problem */
Queue* initial_sieve(uint64_t limit) {
Queue* queue = init_queue();
uint64_t _sqrt = (uint64_t)sqrt(limit);
uint32_t *primes = (uint32_t*)calloc(_sqrt / SIZE + 1, sizeof(uint32_t));
set_bit(primes, 0);
// working with reversed logic, otherwise primes should all me initialized to max uiint64_t
enqueue(queue, 2);
for (uint64_t number = 3; number <= _sqrt; number += 2) {
if (get_bit(primes, number / 2) == 0) {
enqueue(queue, number);
for (uint64_t position = number * number; position <= _sqrt; position += (number * 2)) {
set_bit(primes, position / 2);
}
}
else
set_bit(primes, number / 2);
}
return queue;
}
uint64_t count_primes(uint64_t limit) {
uint64_t start, end, delta;
uint64_t non_primes_counter, initial_size;
uint32_t *current_sieve;
Queue* queue;
queue = initial_sieve(limit);
initial_size = queue->size;
start = delta = (uint64_t)sqrt(limit);
end = 2 * start;
non_primes_counter = 0;
printf("Limits: %llu -> %llu\n", start, end);
while (start < limit) {
Node* prime = queue->first->next; // pass 2 since only odd maps are represented in the sieve
uint64_t count = 0;
current_sieve = (uint32_t*)calloc(delta / SIZE + 1, sizeof(uint32_t));
// memset(current_sieve, 0, sizeof(uint32_t) * delta);
while (prime != NULL) {
uint64_t first_composite = start / prime->data * prime->data;
// calculate the first multiple of the given prime in the interval
if (first_composite < start)
first_composite += prime->data;
if ((first_composite & 1) == 0)
first_composite += prime->data;
// set all the composites of the current prime in the given interval
for (uint64_t number = first_composite; number <= end; number += (prime->data) * 2) {
const uint64_t position = (number - start) / 2;
if (get_bit(current_sieve, position) == 0) {
set_bit(current_sieve, position);
count++;
}
}
// free(current_sieve);
prime = prime->next;
}
non_primes_counter += count;
start += delta;
end += delta;
if (end > limit)
end = limit;
}
uint64_t total = (limit - delta + 1) / 2 - non_primes_counter;
printf("%sTotal composites and initial size: %llu %llu %s\n", KCYN, non_primes_counter, initial_size, RESET);
printf("%sTotal primes: %llu %s\n", KCYN, total, RESET);
return queue->size + (limit - delta + 1) / 2 - non_primes_counter;
}
/* Main */
int main(int argc, char **argv) {
clock_t begin, end;
double time;
if (argc < 2) {
printf("Invalid number of parameters\n");
printf("Program will exit now.\n");
return 0;
}
begin = clock();
printf("%sNumber of primes found up to %s%s: %s%llu.\n%s", KWHT, KCYN, argv[1], KYEL, count_primes(atoll(argv[1])), RESET);
end = clock();
time = (double)(end - begin) / CLOCKS_PER_SEC;
printf("%sTotal time elapsed since the starting of the program: %s%lf seconds.\n%s", KWHT, KYEL, time, RESET);
return 0;
}
谢谢，马库斯
答案 0 :(得分：2)
你需要计算素数的数量，而不是全部找到它们（它们太多了）。这称为Prime-counting function。
在数学中，计数功能是计算小于或等于某个实数x的素数的函数。它用π（x）表示。
有很多方法可以计算出这个功能。使用方法比较查看此Wolfram page。在两分钟内完成这项工作似乎是一项艰巨的任务。
正如评论中提到的，还有一个很棒的answer at math.stackexchange，我认为这会有所帮助。