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?

i guess you are missing the output version , the output says 4 space 9 and you are printing 4 and 9 together .

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.

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.

MISTERM Problem - CodeChef . … a very common question in almost every site .