Missing number in array

Given an array of size n-1 and given that there are numbers from 1 to n with one missing, the missing number is to be found.

Input:

The first line of input contains an integer T denoting the number of test cases.
The first line of each test case is N,size of array.
The second line of each test case contains N-1 input C[i],numbers in array.

Output:

Print the missing number in array.

Constraints:

1 ≤ T ≤ 100
1 ≤ N ≤ 1000
1 ≤ C[i] ≤ 1000

Example:

Input
2
5
1 2 3 5
10
1 2 3 4 5 6 7 8 10

Output
4
9

MY CODE IS AS FOLLOWS:-

#include

using namespace std;

int main() {
int number,size,b[999],d[999],a,c,e;
cin>>number;
for(int i=0;i<number;i++){
cin>>size;
for( a=1;a<=(size-1);a++){cin>>b[a];

    }
    for( c=1;c<=size;c++){
    	d[c]=c;
        
        }
        for(e=1;e<=size;e++){
        	if(d[e]!=b[e]){
        		cout<<d[e];
        		break;
			}
		}
    }
//code
return 0;

}

what is wrong in this?

instead of posting your code post link to your solution.

We need problem link to make sure its not from a on going coding contest, as it had been done in past.

https://www.codechef.com/problems/MISTERM . … a very common question in almost every site .

i guess you are missing the output version , the output says 4 space 9 and you are printing 4 and 9 together .
http://ide.geeksforgeeks.org/eDIlos

You have assumed the array to be in a sorted order. The given array may or may not be in sorted order. If the array isn’t sorted your solution will not work.

Declare an array position of size n which all its element initialized as zero.

While taking input for each input temp do position[temp]=1

At the end check which element in the array is still 0, that will be your answer.

//