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We will use this function to proceed and store the result 

void find_two_largest(int a[], int n, int *largest, int *second_largest);

* When passed an array a of length n, the function will search a for its largest and second-largest elements, storing them in the variables pointed to by largest and second_largest respectively.

寫 find_two_largest() 函數。陣列做為函數參數 找出最大值跟第二大值。

Finish the code below to match the input and output. Pointer is required. 請使用指標。

#include <stdio.h>

void find_two_largest(int a[], int n, int *largest, int *second_largest);

int main(void)
{
	int n, largest, second_largest;
	scanf("%d", &n); //Numbers will be entered
	int a[n];
    //Enter n integers separated by spaces
	for (int i = 0; i < n; i++)
		scanf(" %d", &a[i]);

	find_two_largest(a, n, &largest, &second_largest);

	if (n == 0)
		// Your code here
		//Print No numbers were entered.
	else if (n == 1)
		// Your code here
		//Print Only one number was entered. Largest: 
	else
		// Your code here
		//Print Largest: , Second Largest:

	return 0;
}

void find_two_largest(int a[], int n, int *largest, int *second_largest)
{
	// Your code here (Using pointer to finish)
}

Example input:

0

Example output:

No numbers were entered.

Example input:

1
67

Example output:

Only one number was entered. Largest: 67

Example input:

5
5 6 7 8 9

Example output:

Largest: 9, Second Largest: 8

 

[Exercise] Coding (C) - asked in Chapter 12: Pointers and Arrays
ID: 38251 - Available when: Unlimited - Due to: Unlimited

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#include * * * *
void find_two_largest(int a[], int n, int *largest, int *second_largest);

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#include * *** * ** **

void ** * ** * a[], int n, int *largest, int ** ** *

int main(void)
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#include * ** *****

void ** ** * ** * a[], int n, int *largest, int *second_largest);

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#include * *** *

void * ** ** a[], int n, int *largest, int * * *

int *
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#include ***** ** **

void * * * *** *** a[], int n, int **** ** int * * *

int ***** *
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****** * * * return 0;
}

void * * * ** a[], int n, int * * int ** **
{
***** * * *** // Your code here (Using pointer to finish)
**** * ** * int i=0;
**** ** int j=0;
* **** *** * * ***** = ** ** * a[0];
* * * for (i = 1; i ** ** n; i++) {
** * * ******** * ** (a[i] ** * **
*** **** * * * * **** ** *** *** *** **
* * *** *** *** * * ** * * ***** ** * ** * *** * **** **** * = a[i];
* *** * * * * * * * *
* *** * ** * ** * * * ** * ** (j = 1; j * n; j++)
** ******** * * * ** ** * ** * *** ** ** * * * ** * **** *** ** * **
** ** * ****** * * *** * ** * * ** * **** * *
** ** * * * ** * **** * * *** ** * *** * * * * ** * ** * ****** * = a[j];
* * * ** ** * * ***** *** * * ** *
** ** * ** * }
}
answered by (-329 points)
0 0
Case 0: Correct output
Case 1: Wrong output
Case 2: Correct output
Case 3: Correct output
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