A Amr and Music（ Codeforces Round #287 (Div. 2)）

A. Amr and Music

time limit per test

memory limit per test

input

output

Amr is a young coder who likes music a lot. He always wanted to learn how to play music but he was busy coding so he got an idea.

n instruments, it takes ai days to learn i-th instrument. Being busy, Amr dedicated k

Amr asked for your help to distribute his free days between instruments so that he can achieve his goal.

Input

n, k (1 ≤ n ≤ 100, 0 ≤ k ≤ 10 000), the number of instruments and number of days respectively.

n integers ai (1 ≤ ai), representing number of days required to learn the i-th instrument.

Output

m

m

if there are multiple optimal solutions output any. It is not necessary to use all days for studying.

Sample test(s)

input

4 10 4 3 1 2

output

4 1 2 3 4

input

5 6 4 3 1 1 2

output

3 1 3 4

input

1 3 4

output

0

Note

4

{2, 3, 5} or {3, 4, 5}.

In the third test Amr doesn't have enough time to learn the only presented instrument.

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>

using namespace std;

int n,m;
struct node
{
    int x,y;
}a[14100];

int cmp(const void *a,const void *b)
{
    struct node *aa,*bb;
    aa=(struct node *)a;
    bb=(struct node *)b;
        return aa->x-bb->x;
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i].x);
            a[i].y = i+1;
        }
        qsort(a,n,sizeof(a[0]),cmp);
        int b[10010];
        int t = 0;
         memset(b,0,sizeof(b));
        for(int i=0;i<n;i++)
        {
            if(m>=a[i].x)
            {
                m = m-a[i].x;
                b[t++] = a[i].y;
            }
            else
            {
                break;
            }
        }
       sort(b,b+t);
        printf("%d\n",t);
        for(int i=0;i<t;i++)
        {
            if(i == t-1)
            {
                printf("%d\n",b[i]);
            }
            else
            {
                printf("%d ",b[i]);
            }
        }
    }
    return 0;
}