我正在尝试查找给定数N的素数并将其返回到链接列表中。查找素数没有问题,但是在链接列表中返回它们却有问题...我没有得到我在运行代码时出错,但我仅获得第一个素因数作为输出,而无法得到余数,例如,如果N等于72,我将得到2作为输出,但无法保留其余因数
#include <stdio.h>
#include <stdlib.h>
//This is my structure
typedef struct SinglyLinkedListItem
{
int data;
struct SinglyLinkedListItem*next;
}SLLI;
//This is my function to find the prime factor and return in a linked list
SLLI*PrimeFactor(SLLI*prime,int N)
{
SLLI*pList=NULL;
int i,j,isPrime;
for (i = 2; i <= N; i++)
{
if(N % i == 0)
{
isPrime = 1;
for (j = 2; j <= i/2; j++)
{
if(i % j == 0)
{
isPrime = 0;
break;
}
}
//Most probably problem is after this part of the code
if(isPrime == 1)
{
//Adding first factor to the list
if(pList==NULL)
{
pList=malloc(sizeof(SLLI));
pList->data=i;
pList->next=NULL;
prime= pList;
}
//Trying to add rest of them but can't
else
{
pList->next = malloc(sizeof(SLLI));
pList->next->data = i;
pList->next->next = NULL;
pList = pList->next;
}
}
}
}
return prime;
}
void Printlist(SLLI*pHead)
{
SLLI*temp=pHead;
while(temp!=NULL)
{
printf("%d\t",temp->data);
temp=temp->next;
}
}
int main()
{
SLLI*pHead=NULL;
pHead=PrimeFactor(pHead,72);
Printlist(pHead);
}
[我怀疑您只是没有正确地打印它们,由于您没有包含该代码,因此很难说,但是如果您这样添加main
:
int main(void) {
SLLI * x = PrimeFactor(NULL, 72);
while (x != NULL) {
printf("%d\n", x->data);
x = x->next;
}
return 0;
}
然后您将同时获得2
和 3
,这是72
的唯一主要因素:72 = 2332 (8 x 9)
。
类似地,120
为您提供2
,3
和5
:120 = 233151 (8 x 3 x 5)
。
要考虑的其他几点:
我不确定为什么将prime
传递给函数,因为无论如何都会覆盖它。您应该从函数定义中删除它,而只使用一个局部变量来保存信息。
当检查素数时,您不必将值取一半,只需取其平方根即可。因此,更好的循环是:for (j = 2; j * j <= i; j++)
。
偶数[[更好检查是要认识到,除2
和3
之外,每个素数均采用6n + 1
或6n + 5
的形式。那是因为:
6n + 0 = 6n
,六的倍数;]6n + 2 = 2(3n + 1)
,两个的倍数;]6n + 3 = 3(2n + 1)
,三的倍数;]6n + 4 = 2(3n + 2)
,两个的倍数;]6n + 1
和6n + 5
作为候选项(不是每个one
6(4) + 1 = 25
),但是素数can// Function for checking primality.
int isPrime(int number) {
// Special cases.
if ((number == 2) || (number == 3)) return 1;
if ((number % 2 == 0) || (number % 3 == 0)) return 0;
if (number < 5) return 0;
// Efficient selection of candidate primes, starting at 5, adding
// 2 and 4 alternately: 5, 7, 11, 13, 17, 19, 21, 25, 27, ...
for (
int candidate = 5, add = 2;
candidate * candidate <= number;
candidate += add, add = 6 - add
) {
if (number % candidate == 0) {
return 0;
}
}
return 1;
}
使用该功能,并添加一些额外的功能以转储和释放列表,并提供测试工具main
,将提供以下完整程序:
#include <stdio.h>
#include <stdlib.h>
typedef struct sListItem
{
int data;
struct sListItem *next;
} ListItem;
// Function for checking primality.
int isPrime(int number) {
// Special cases.
if ((number == 2) || (number == 3)) return 1;
if ((number % 2 == 0) || (number % 3 == 0)) return 0;
if (number < 5) return 0;
// Efficient selection of candidate primes, starting at 5, adding
// 2 and 4 alternately: 5, 7, 11, 13, 17, 19, 21, 25, 27, ...
for (int candidate = 5, add = 2; candidate * candidate <= number; candidate += add, add = 6 - add)
if (number % candidate == 0)
return 0;
return 1;
}
// Function for returning list of prime factors.
ListItem *primeFactorList(int number) {
ListItem *retVal = NULL;
ListItem *lastItem;
// Analyse possible factors, up to half of number.
for (int divisor = 2; divisor <= number / 2 + 1; divisor++) {
if ((number % divisor == 0) && isPrime(divisor)) {
if (retVal == NULL) {
// Adding first item to list.
retVal = lastItem = malloc(sizeof(ListItem));
lastItem->data = divisor;
lastItem->next = NULL;
} else {
// Adding subsequent items to list.
lastItem->next = malloc(sizeof(ListItem));
lastItem = lastItem->next;
lastItem->data = divisor;
lastItem->next = NULL;
}
}
}
return retVal;
}
// Dump a list.
void dumpList(int value, ListItem *head) {
printf("%d:", value);
while (head != NULL) {
printf(" -> %d", head->data);
head = head->next;
}
putchar('\n');
}
// Free a list.
void freeList(ListItem *head) {
while (head != NULL) {
ListItem *toDelete = head;
head = head->next;
free(toDelete);
}
}
// Test program.
int main(int argc, char *argv[]) {
static int data[] = { 10, 24, 72, 120, 125, -1 };
for (int *ptr = &(data[0]); *ptr >= 0; ptr++) {
ListItem *list = primeFactorList(*ptr);
dumpList(*ptr, list);
freeList(list);
}
return 0;
}
并显示测试工具的结果进行编译和运行(在右侧添加了我的注释,如果需要更多测试,请随时向data
数组添加任何额外的值):
10: -> 2 -> 5 2 x 5 24: -> 2 -> 3 2^3 x 3 72: -> 2 -> 3 2^3 x 3^2 120: -> 2 -> 3 -> 5 2^3 x 3 x 5 125: -> 5 5^3