1、一堆硬幣，一個(gè)機(jī)器人，如果是反的就翻正，如果是正的就拋擲一次，無窮多次后，求正反的比例

解答：是不是題目不完整啊，我算的是3：1

2、一個(gè)汽車公司的產(chǎn)品，甲廠占40%，乙廠占60%，甲的次品率是1%，乙的次品率是2%，現(xiàn)在抽出一件汽車時(shí)次品，問是甲生產(chǎn)的可能性

解答：典型的貝葉斯公式，p(甲|廢品) = p(甲 && 廢品) / p(廢品) = (0.4 × 0.01) /(0.4 × 0.01 + 0.6 × 0.02) = 0.25

3、k鏈表翻轉(zhuǎn)。給出一個(gè)鏈表和一個(gè)數(shù)k，比如鏈表1→2→3→4→5→6，k=2，則翻轉(zhuǎn)后2→1→4→3→6→5，若k=3,翻轉(zhuǎn)后3→2→1→6→5→4，若k=4，翻轉(zhuǎn)后4→3→2→1→5→6，用程序?qū)崿F(xiàn)

非遞歸可運(yùn)行代碼：

#include

#include

#include

typedef struct node {

struct node *next;

int data;

} node;

void createList(node **head, int data)

{

node *pre, *cur, *new;

pre = NULL;

cur = *head;

while (cur != NULL) {

pre = cur;

cur = cur->next;

}

new = (node *)malloc(sizeof(node));

new->data = data;

new->next = cur;

if (pre == NULL)

*head = new;

else

pre->next = new;

}

void printLink(node *head)

{

while (head->next != NULL) {

printf("%d ", head->data);

head = head->next;

}

printf("%d ", head->data);

}

int linkLen(node *head)

{

int len = 0;

while (head != NULL) {

len ++;

head = head->next;

}

return len;

}

node* reverseK(node *head, int k)

{

int i, len, time, now;

len = linkLen(head);

if (len < k) {

return head;

} else {

time = len / k;

}

node *newhead, *prev, *next, *old, *tail;

for (now = 0, tail = NULL; now < time; now ++) {

old = head;

for (i = 0, prev = NULL; i < k; i ++) {

next = head->next;

head->next = prev;

prev = head;

head = next;

}

if (now == 0) {

newhead = prev;

}

old->next = head;

if (tail != NULL) {

tail->next = prev;

}

tail = old;

}

if (head != NULL) {

tail->next = head;

}

return newhead;

}

int main(void)

{

int i, n, k, data;

node *head, *newhead;

while (scanf("%d %d", &n, &k) != EOF) {

for (i = 0, head = NULL; i < n; i ++) {

scanf("%d", &data);

createList(&head, data);

}

printLink(head);

newhead = reverseK(head, k);

printLink(newhead);

}

return 0;

}