Help me fix this code: Finding the missing number in the array

[polkodot]

I'm trying to write a program where in you input 9 numbers from 1-10, then it determines the missing number. Here's my code. It has a lot of errors. Help me do the right structure. Maybe after that, I'll improve it by setting conditions like no repetition/ 0< number <10. Just help me to fix this first:

#include <iostream>
using namespace std;

int main()
{
	int d[]={1,2,3,4,5,6,1,7,9, 10};
	int i=0,j=0,c=0,missing=0;

for (i = 0; i < 9; i++)
{
    cout << "input: ";
    cin >> d[i];
	c = 0;
	for (j = 0; j<10; j++)
	{
		if (i == d[j])
		{
			c = i;
		}
	}
	if (i==d[j])
	{
		missing = j;

	}
}
	cout << "missing is " << missing << endl;
	return 0;
}

I'm so new to C++. It always prints out the number 10 even though I typed it already. I really have no idea how to determine the missing one.

[whodgson]

There are 10 numbers between 1 and 10 inclusive and there are 10
integers in d[]. Your two loop conditions are 8 and 9. Other than those observations I have no idea what you are doing.
*"It always prints out the number 10"
this is because when j<10 or 9 (and finishes iterating)it has reached the last array element which contains 10. Remember j starts counting at 0.

[wholetthedogout]

Can't this be an easiest approach

1) Sort the given sequence
2) fix the no. with the difference of two consecutive no. as > 1; duplicates can be filtered out with the difference as zero.

Hope this help.

[whodgson]

Or you could:
1)Inside a loop iterate through the array and at each element compare loop i with array[i].If they are equal (==) continue.

2)If (array[i]!= loop (i))cout<<i<<" ";//not in array so print