# Print elements that can be added to form a given sum

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Input number of elements user want to enters and than input the total value user want to calculate from the given list of elements.

Input : N=5
Enter any 5 values : 3 1 6 5 7
Enter sum you want to check : 10
Output : 3 1 6

## Algorithm

START
STEP1-> Take values from the user
STEP2-> Take the sum a user want to check in the set.
STEP3-> For i = 0; i < n; i++
STEP4-> Check If sum - *(ptr+i) >= 0 then,
STEP4.1-> sum -= *(ptr+i);
STEP4.2-> Print the value of *(ptr+i)
END If
END For
STOP

## Example

#include <stdio.h>
int main(int argc, char const *argv[]){
int *ptr, n, i, sum;
printf("Enter number of digits you want to enter");
scanf("%d", &n);
ptr = (int*)malloc(sizeof(int)*n); //Dynamically allocating the memory of int
type
printf("Enter %d elements", n);
for(i = 0; i < n; i++) {
scanf("%d", (ptr+i)); //Inputting the value in dynamically
//allocated array
}
printf("Enter the sum you want to check");
scanf("%d", &sum);
for ( i = 0; i < n; i++) {
if(sum - *(ptr+i) >= 0) { //Checking the values which can be added to form the sum
X
sum -= *(ptr+i); //Updating the value of sum
printf("%d ", *(ptr+i)); //Printing the Values which can be summed up to form sum
}
}
return 0;
}

## Output

If we run the above program then it will generate the following output

Enter number of digits you want to enter
5
Enter 5 elements
3
1
6
5
7
Enter the sum you want to check
10
3 1 6
Updated on 30-Jul-2019 22:30:26