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

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* * ** *******

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#include * * ** *

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

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#include * ****

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* * * * ** *** {// Your code here
* * ** * * **** * * * * * ** one number was *** *** *** * *** *
*** *** **** * * * }
*** else
* * *** **** * * *** {// Your code here
* * ***** ** * * *** * * * * ***** **** %d, Second * * ** *** ** * ** **
* *** ** * *** ** ** * }
*** * 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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