codeforces 507A Amr and Music

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A. Amr and Music

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard 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.

Amr has n instruments, it takes ai days to learn i-th instrument. Being busy, Amr dedicated k days to learn how to play the maximum possible number of instruments.

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

Input

The first line contains two numbers n, k (1 ≤ n ≤ 100, 0 ≤ k ≤ 10 000), the number of instruments and number of days respectively.

The second line contains n integers ai (1 ≤ ai ≤ 100), representing number of days required to learn the i-th instrument.

Output

In the first line output one integer m representing the maximum number of instruments Amr can learn.

In the second line output m space-separated integers: the indices of instruments to be learnt. You may output indices in any order.

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

Examples

input

4 104 3 1 2

output

41 2 3 4

input

5 64 3 1 1 2

output

31 3 4

input

1 34

output

0

Note

In the first test Amr can learn all 4 instruments.

In the second test other possible solutions are: {2, 3, 5} or {3, 4, 5}.

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

题意懂了就会了。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;struct node{    int x,y;} a[110];int comp(node x,node y){    return x.x<y.x;}int main(){    int n,k,b[110],w=0;    scanf("%d%d",&n,&k);    for(int i=0; i<n; i++)    {        scanf("%d",&a[i].x);        a[i].y=i;    }    sort(a,a+n,comp);    for(int i=0;i<n;i++)    {        if(k>=a[i].x)        {            k-=a[i].x;            b[w++]=a[i].y;        }    }    printf("%d\n",w);    for(int j=0;j<w;j++)    {        if(j==0)            printf("%d",b[j]+1);        else            printf(" %d",b[j]+1);    }    if(w)        printf("\n");    return 0;}