2016 Winter Training Day #1_F题_codefcrces 507A(贪心)_amr is a young coder who likes music a lot. he alw-优快云博客

本文概述了AI音视频处理领域的关键技术，包括视频分割、语义识别、自动驾驶、AR、SLAM等，并探讨了其在实际应用中的作用。

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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.

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

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：给出n个乐器和k天时间。每个乐器需要ai天时间学会，求k天内学会的最多的乐器数量及其序号。

思路：简单贪心，将n个乐器按需要的时间升序排列，先学需要的时间短的，最后输出该原来序号（用结构体保存起来）。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>

using namespace std;

const int INF = 0x7fffffff;

struct Node
{
    int cost;
    int no;
};
Node a[120];
bool cmp(Node a, Node b)
{
    return a.cost < b.cost;
}

int main()
{

    int n, k;
    cin >> n >> k;
    int ans = 0;
    int sum = 0;
    for(int i = 0; i < n; ++i)
    {
        cin >> a[i].cost;
        a[i].no = i;
    }
    sort(a, a+n, cmp);
    for(int i = 0; i < n; ++i)
    {
        if(sum+a[i].cost <= k)
        {
            sum += a[i].cost;
            ans ++;
        }
        else
            break;
    }
    cout << ans << endl;
    for(int i = 0; i < ans; ++i)
        cout << a[i].no+1 << " ";
    cout << endl;
    return 0;
}