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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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Hidden content!
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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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* ** * ***

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** * * **** *** * *

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Hidden content!
#include *** * ** * *

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

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void find_two_largest(int a[], int n, int *largest, int *second_largest)
{
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}
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Hidden content!
#include * *** * *****

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

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

int *
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* * * ** * 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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