Mark and Toys

                            Solution in C :

In  C  :





#include<stdio.h>
void quicksort(int x[100000],int first,int last){
    int pivot,j,temp,i;
 
     if(first<last){
         pivot=first;
         i=first;
         j=last;
 
         while(i<j){
             while(x[i]<=x[pivot]&&i<last)
                 i++;
             while(x[j]>x[pivot])
                 j--;
             if(i<j){
                 temp=x[i];
                  x[i]=x[j];
                  x[j]=temp;
             }
         }
 
         temp=x[pivot];
         x[pivot]=x[j];
         x[j]=temp;
         quicksort(x,first,j-1);
         quicksort(x,j+1,last);
 
    }}

int main()
{
int n,k,i,avail=0,count=0;
scanf("%d",&n);
scanf("%d",&k);
int cost[n];
for(i=0;i<n;i++)
scanf("%d",&cost[i]); 
quicksort(cost,0,n-1);
while(avail<=k)
{
avail+=cost[count]; 
count++;
}  
printf("%d\n",count-1);    
return 0;    }
                        

                        Solution in C++ :

In  C++ :






#include <string>
#include <vector>
#include <map>
#include <list>
#include <iterator>
#include <set>
#include <queue>
#include <iostream>
#include <sstream>
#include <stack>
#include <deque>
#include <cmath>
#include <memory.h>
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <algorithm>
#include <utility> 
using namespace std;
 
#define FOR(i, a, b) for(int i = (a); i < (b); ++i)
#define RFOR(i, b, a) for(int i = (b) - 1; i >= (a); --i)
#define REP(i, N) FOR(i, 0, N)
#define RREP(i, N) RFOR(i, N, 0)
 
#define ALL(V) V.begin(), V.end()
#define SZ(V) (int)V.size()
#define PB push_back
#define MP make_pair
#define Pi 3.14159265358979

typedef long long Int;
typedef unsigned long long UInt;
typedef vector <int> VI;
typedef pair <int, int> PII;

int a[1<<17], n, k;

clock_t startTime;

int main()
{
//	freopen("in.txt", "r", stdin);
//	freopen("out.txt", "w", stdout);
	
	startTime = clock();
	
	scanf("%d%d",&n,&k);
	
	REP(i,n)
	{
		scanf("%d",&a[i]);
	}
	
	sort(a,a+n);
	
	Int sum = 0;
	int res = 0;
	
	while (sum <= k)
	{
		sum += a[res];
		
		if (sum > k)
			break;
		
		++res;
	}
	
	printf("%d\n", res);
	
//	fprintf(stderr, "Used time: %f", (clock() - startTime) / (CLOCKS_PER_SEC*1.0));
	
	return 0;
}
                    

                        Solution in Java :

In  Java :








import java.io.*;
import java.util.*;

public class Solution{		
		
		private BufferedReader in;	
		private StringTokenizer st;
		private PrintWriter out;
		
		
		
		
		void solve() throws IOException{
			
		
			int n = nextInt();
			int k = nextInt();
			int []x = new int[n];
			for (int i = 0; i < x.length; i++) {
				x[i] = nextInt();
			}
			Arrays.sort(x);
			long sum = 0;
			int ans = 0;
			for (int i = 0; i < x.length; i++) {
				sum += x[i];
				if(sum <= k){
					ans++;
				}
				else
					break;
			}
			out.println(ans);
					
		}
			

		Solution() throws IOException {
			in = new BufferedReader(new InputStreamReader(System.in));	
			out = new PrintWriter(System.out);
			eat("");
			solve();	
			out.close();
		}

		private void eat(String str) {
			st = new StringTokenizer(str);
		}

		String next() throws IOException {
			while (!st.hasMoreTokens()) {
				String line = in.readLine();				
				if (line == null) {					
					return null;
				}
				eat(line);
			}
			return st.nextToken();
		}

		int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public static void main(String[] args) throws IOException {
			new Solution();
		}

		int gcd(int a,int b){
			if(b>a) return gcd(b,a);
			if(b==0) return a;
			return gcd(b,a%b);
		}

}
                    

                        Solution in Python : 
                            
In  Python3 :





x = input ()
parts = x.split ()
part1 = int (parts [0])
part2 = int (parts [1])
y = input ()
arr = []
y = y.split ()
for i in range (0, part1):
	arr.append (int (y [i]))
arr.sort ()
i = 0
j = 0
while j <part1 and part2> arr [j]:
	part2- = arr [j]
	j + = 1
	i + = 1
print (i)