Memorization of a recursive algorithm [closed] - c

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Question: Given an amount m and an array of numbers b[] print "Yes" if m can be formed by addition of any number of elements from the array without repeating them.
This is what I've come up with; how can I add a memoization matrix into it?
#include<stdio.h>
int check(int b[],int n,int m){
if(m==0){
return 1;
}
if(m<0){
return 0;
}
if(n<=0 && m>0){
return 0;
}
return check(b,n-1,m-b[n-1]) + check(b,n-1,m);
}
int main(){
int t;
scanf("%d",&t);
while(t--){
int n,m;
scanf("%d",&n);
scanf("%d",&m);
int b[n];
for(int i=0;i<n;i++){
scanf("%d",&b[i]);
}
if(check(b,n,m)>0){
printf("Yes\n");
}else{
printf("No\n");
}
}
return 0;
}
Can any one help me put a memorization matrix into it ?

Answer 1

The key to the memoization cache consists of the first n elements of b, plus m. Form a string from these, and use it as a key to a hash table.
...But what's the point? With n decreasing in each recursion, there are only two situations where memoization will help:
When a non-recursive call check is made for exactly the same list of numbers and m as an earlier call to check.
When m-b[n-1] == m, which is to say when b[n-1] == 0.
The first situation is unlikely (and decreasingly so as the size of the list increases, which is when memoization would help the most), and the second is easy to avoid. Just replace
return check(b,n-1,m-b[n-1]) + check(b,n-1,m);
with
return b[n-1] == 0
? check(b,n-1,m) * 2
: check(b,n-1,m) + check(b,n-1,m-b[n-1]);