gistfile1.txt

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int cases, note_count, required_money;
//cases - no of cases
//note_count - no of notes
//required money - money asked by the monster

vector<int> notes(20); //stores the values of the notes (there can be at most 20 notes; so let's reserve space for them - insignificant optimization)

bool issubset(int spos, int current_notes)
{
	for (int i = spos; i >= 0; i--)
	{
		if (notes[i] + current_notes > required_money) continue;
		int new_current_notes = current_notes + notes[i];

		if (new_current_notes == required_money) return true;
		if (issubset(i - 1, new_current_notes)) return true;
	}
	return false;
}
int main()
{
	cin >> cases;
	while (cases--)
	{
		bool success = false;
		cin >> note_count >> required_money;

		notes.clear(); //empty the notes vector
		for (int i = 0, note; i < note_count; i++)
		{			
			cin >> note;
			if(note < required_money) //if the value of note is more than the required money, it obviously won't help in forming a subset; these are useless notes and let's not add them to the note pool
				notes.push_back(note);
			else if (note == required_money) //if there is a note whose value is equal to the required money, then we hit a jackpot; let's finish the case without further computation
			{				
				success = true;
				break;
			}
		}
		if (success)
		{
			cout << "Yes" << endl;
			continue; //goes to next case
		}
		note_count = notes.size(); //we might have removed some notes; so let's update the note_count with the correct number of notes in notes vector
		sort(notes.begin(), notes.end()); //sort the notes in increasing order

		int minpos = 0; //1 3 5 7 ... if the required number is 20, there is no combination of notes of the first 3 notes (1, 3, 5) which can give 20; minpos is the index of the next highest note after the notes in the set (in this case 7)
		for (int count = 0; minpos < note_count; minpos++)
		{
			count += notes[minpos]; //add the next note to the sum
			if (count > required_money) break; //the sum exceeds the required money so we need to take the note at this location seriously
		}
		for (int i = note_count - 1; i >= minpos; i--)
		{
			if (issubset(i - 1, notes[i]))
			{
				success = true;
				break;
			}
		}
		if(success) cout << "Yes" << endl;
		else cout << "No" << endl;
	}
	return 0;
}
/*
ALGORITHM
-------------------
let notes be: 1 2 4 6 12 14 23 29 43
required money: 26

29, 43 are greater than 26; so they are useless; we will treat the situation as if they weren't even there

so the useful notes are: 1 2 4 6 12 14 23

LABLE1:
we start from the last number: 23
23 + 14 exceeds 26; so any combination having these two notes are of no help
23 + 12 exceeds 26 too; so any combo with these notes is of no use
23 + 6 and 23 + 4 exceed 26 too; so this combo is useless as well

23 + 2 is less than 26; so we need to check from here

let us combine the two notes 23 and 2; we get 25; having two notes of 23 and 2 is as good as having a note of 25

LABLE2: 
now let us jump to LABLE1 and start with the number 25 and check notes smaller than 2
25 + 1 = 26!!!! this works!!

Suppose 25 + 1 hadn't worked, we would try the next number after 2; try with 23 + 1 + x combo 
(in this case there can't be any x becaz 1 is the smallest)

so the next change can happen with 23
now we know that no combination having 23 can give the required money

so let's go to next number: 14
GOTO LABLE1 with next number as 14
*/