以下功能是带有问题和答案的游戏的一部分。我对问题随机化的问题有一个问题 - 一些问题出现不止一次,第一个也出现在每个级别,这不应该发生。请帮忙!
typedef struct
{
char question[300];
char date[30], author[30], ansr1[80], ansr2[80], ansr3[80], ansr4[80];
int id, correctAnsr, level;
} game;
typedef struct Node
{
game data;
struct Node* next;
} node;
void printRandom1(node *head)
{
if (isEmpty(head))
return;
srand(time(NULL));
node *result = head->data.question;
node *current = head;
int n;
for (n = 17; current != NULL; n++)
{
if (rand() % n == 0 && current->data.level == 0)
result = current->data.question;
current = current->next;
}
printf("%s\n", result);
int i, ans;
printf("1.-%s\n", result->data.ansr1);
printf("2.-%s\n", result->data.ansr2);
printf("3.-%s\n", result->data.ansr3);
printf("4.-%s\n", result->data.ansr4);
printf("Enter the correct answer: ");
scanf("%d", &ans);
if (ans == result->data.correctAnsr)
printf("CORRECT ANSWER!\n");
else
{
printf("WRONG ANSWER\n");
printf("GAME OVER");
printf("The correct answer is: %d\n", result->data.correctAnsr);
return menu();
exit(1);
}
}
答案 0 :(得分:1)
广告“某些问题不止一次出现”:这是因为您无论如何都不会跟踪使用的问题 - 您的随机选择方法始终从所有问题的列表中选择(无论他们是否已经有人问过。)
广告“首先也出现在每个级别”:我的赌注是你的随机选择方法(这很奇怪)不能保证选择一个问题(即result = current->data.question
部分可能不会以很高的概率执行)。在这种情况下保留result
的初始值(这恰好是第一个问题)。
以下是您的代码的修改版本。一些评论:
添加链接列表中的问题数量。随机选择需要选择具有相同概率的答案(正确 - 存在一些可忽略的偏差,但在这里可能不重要)
在新的链接列表中跟踪使用的答案
没有实现级别逻辑。您可能希望从游戏开头的链接列表中删除不适当级别的问题
current
变量是一个指向指针的指针,它简化了取消链接过程(你不需要以这种方式保留前一个入口指针)
代码:
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
typedef struct
{
char question[300];
char date[30], author[30], ansr1[80], ansr2[80], ansr3[80], ansr4[80];
int id, correctAnsr, level;
} question;
typedef struct questionListNode {
question data;
struct questionListNode* next;
} questionListNode;
typedef struct {
questionListNode* headAvailable; // Your former head variable
questionListNode* headUsed; // Initialize this to NULL
int numberOfAvailableNodes; // Number of nodes in headAvailable
int numberOfCorrectAnswers; // Number of nodes in headUsed
} game;
void printRandom1(game* currentGame)
{
if (currentGame->headAvailable == NULL || currentGame->numberOfAvailableNodes <= 0) {
printf("No more questions, you've won!\n");
exit(1);
}
srand(time(NULL)); // Consider moving this to the start of main()
int chosenIndex = rand() % currentGame->numberOfAvailableNodes;
questionListNode** current = &(currentGame->headAvailable);
while ((chosenIndex > 0) && (*current != NULL)) { // the second check is for safety
current = &((*current)->next);
chosenIndex--;
}
questionListNode* currentQuestion = (*current);
if (currentQuestion == NULL) {
printf("Internal error: available count mismatch!\n");
exit(1);
}
printf("%s\n", currentQuestion->data.question);
int i, ans;
printf("1.-%s\n", currentQuestion->data.ansr1);
printf("2.-%s\n", currentQuestion->data.ansr2);
printf("3.-%s\n", currentQuestion->data.ansr3);
printf("4.-%s\n", currentQuestion->data.ansr4);
printf("Enter the correct answer: ");
scanf("%d", &ans);
if (ans != currentQuestion->data.correctAnsr) {
printf("WRONG ANSWER\n");
printf("GAME OVER\n");
printf("The correct answer is: %d\n", currentQuestion->data.correctAnsr);
exit(1);
}
printf("CORRECT ANSWER!\n");
// Remove currentQuestion from the available list
(*current) = currentQuestion->next;
// Put currentQuestion into used list
currentQuestion->next = currentGame->headUsed;
currentGame->headUsed = currentQuestion;
// Update counters
currentGame->numberOfAvailableNodes--;
currentGame->numberOfCorrectAnswers++;
}
int main(int c, char** t)
{
game g;
g.headAvailable = NULL;
g.headUsed = NULL;
g.numberOfAvailableNodes = 0;
g.numberOfCorrectAnswers = 0;
questionListNode q1 = { { "Question 1", "", "", "A1*", "B1", "C1", "D1", 1, 1, 0 }, NULL };
questionListNode q2 = { { "Question 2", "", "", "A2", "B2*", "C2", "D2", 2, 2, 0 }, &q1 };
questionListNode q3 = { { "Question 3", "", "", "A3", "B3*", "C3", "D3", 3, 2, 0 }, &q2 };
g.headAvailable = &q3;
g.numberOfAvailableNodes = 3;
while (1)
printRandom1(&g);
}
一些额外的(随机)笔记:
我不确定link-list是此任务的最佳数据结构
考虑使用一些前缀命名您的typedef(例如t_game
,t_node
)
如果您想重新开始游戏(而不是exit()
),则需要将两个链接列表重新加入并重置计数器
免责声明:我没有花太多时间查看代码,所以请以此为例......