Could anyone help me to get the sum of an array to a fixed value

A=[1 2 3 4;
 5 6 7 8]
how to get the sum of A to be fixed to a 
value of 20 such that all the values in A needs
 to be  changed according to it.

[Kevin Chng]

sum(A)

Will give you the sum. However, it need some programming logic if you want the values within A change according to a fix value of A which is 20.

How to change? It need logic. for example, if sum(A) is more than 20,

which value in A must be reduced so that the sum is 20?

[madhan ravi]

Give an example of your desired output

[jaah navi]

There is no fixed logic A=[1 2 3 4; 4 3 2 1] When I sum the above A it gives 20 the value in A needs to be changed such that the sum of A needs to be 20. the values in A can be negative,or it can be repeated more than once,twice,thrice and so on.

Answer 1

[KSSV]

A=[1 2 3 4;
 5 6 7 8] ;

A = A(:) ;
iwant = cell([],1) ;
count = 0 ;
for i = 1:length(A)
    B = nchoosek(A,i) ;
    thesum = sum(B,2) ;
    idx = thesum==20 ;
    if any(idx)
        count = count+1 ;
        iwant{count} = B(idx,:) ;
    end
end
iwant

[jaah navi]

What i actually need is the sum of A should be 20 and the values in A needs to be changed according to it,but when i run the above code it gives 36.could you please help me on this.

Answer 2

[Bruno Luong]

"There is no fixed logic"

OK that's easy then

A(:) = 0;
A(1) = 20;

[Kevin Chng]

Best Solution.

[jaah navi]

if i use the above command i am getting value only in the first place and the rest of the place is 0. But i need to have values in places where A already has values, provided no fixed logic is that the values of A can be changed .

[Kevin Chng]

How about

A(:)=1;
A(1) = 20-sum(A(2:end));

provided number of element in A lesser than 20.

Answer 3

[Bruno Luong]

Let's be more serious you can do many thing like shifting

A = A - sum(A) + 20/size(A,1);

or scaling

A = 20 * A ./ sum(A);

or both