我有這個代碼，它從用戶輸入的數字中找到數字的出現。我的問題是如何首先由用戶輸入數字，然后讓用戶在序列中找到他的號碼。如果這個問題很愚蠢，我很抱歉，但我只是一個試圖自學編程的初學者:) 非常感謝您的幫助！
這是我當前的代碼：
int main()
{
int num, remainder, k, count = 0;
printf("Which number do you want to find: \n");
scanf("%d", &k);
while(1)
{
printf("Enter numbers(end by pressing 0): ");
scanf("%d", &num);
remainder=num;
if(remainder==k)count ;
if(remainder==0) break;
}
printf("\Number %d occurs %d times.\n", k, count);
getch();
return 0;
}
這是我嘗試過的：
int main()
{
int num, remainder, k, count = 0;
while(1)
{
printf("Enter numbers(end by pressing 0): ");
scanf("%d", &num);
remainder=num;
if(remainder==k)count ;
if(remainder==0) break;
}
printf("Which number do you want to find: \n");
scanf("%d", &k);
printf("\Number %d occurs %d times.\n", k, count);
getch();
return 0;
uj5u.com熱心網友回復：
如果您想先讀取數字串列，則需要使用變數以某種方式將其存盤在記憶體中，然后才能在該串列中找到數字。
這可以通過使用arraysC 中的概念來完成。
通常，陣列存盤類似型別資料的明確集合/串列。例如，一個陣列int存盤一個整數集合。
你可以在這里閱讀更多關于它們的資訊，甚至你也可以谷歌搜索。
來到您的代碼，您可以像這樣在陣列中宣告和存盤資料，
int numbers[10]; // stores 10 integers, size 10 is static here.
int remainder, k, count = 0;
printf("Enter 10 digits to store : ");
for(int i = 0; i < 10; i ) { // this is a for-loop that executes below block 10 times
scanf("%d", &numbers[i]); // read a number to location i
}
printf("Which number do you want to find: \n");
scanf("%d", &k);
// now use for-loop to iterate over the numbers array(while-loop can also be used)
for(int i = 0; i < 10; i ) {
remainder=numbers[i];
if(remainder==k)count ; // you can also directly compare numbers[i] to k
// below line is not needed since the loop stops at 10th iteration
// if(remainder==0) break;
}
printf("\Number %d occurs %d times.\n", k, count);
筆記：
上面的代碼僅在用戶想要給出 10 個數字時才有效。如果用戶需要提供可變數量的數字來查找k，那么您必須從用戶讀取陣列的大小并創建該大小的動態陣列。參考這個
uj5u.com熱心網友回復：
除了我的答案，您可以放心地忽略所有答案。:)
對于您的任務，您需要稱為單鏈表的資料結構。理論上它沒有限制，您可以輸入任何數字序列，直到輸入 0。
這是一個演示程式。
#include <stdio.h>
#include <stdlib.h>
struct node
{
int number;
struct node *next;
};
struct sequence
{
struct node *head;
struct node *tail;
};
int append( struct sequence *seq, int number )
{
struct node *new_node = malloc( sizeof( *new_node ) );
int success = new_node != NULL;
if( success )
{
new_node->number =number;
new_node->next = NULL;
if ( seq->head == NULL )
{
seq->head = new_node;
}
else
{
seq->tail->next = new_node;
}
seq->tail = new_node;
}
return success;
}
size_t count( const struct sequence *seq, int number )
{
size_t cnt = 0;
for ( const struct node *current = seq->head; current != NULL; current = current->next )
{
if ( current->number == number ) cnt;
}
return cnt;
}
void clear( struct sequence *seq )
{
while ( seq->head != NULL )
{
struct node *current = seq->head;
seq->head = seq->head->next;
free( current );
}
seq->tail = NULL;
}
int main(void)
{
struct sequence seq = { .head = NULL, .tail = NULL };
printf( "Enter numbers (end by entering 0): " );
int number;
while ( scanf( "%d", &number ) == 1 && number != 0 && append( &seq, number ) );
number = 0;
printf( "Which number do you want to find: " );
scanf( "%d", &number );
printf( "\nNumber %d occurs %zu times.\n", number, count( &seq, number ) );
clear( &seq );
return 0;
}
它的輸出可能看起來像
Enter numbers (end by entering 0): 1 2 3 4 5 6 7 8 9 1 8 7 6 5 4 3 2 1 0
Which number do you want to find: 1
Number 1 occurs 3 times.
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