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

    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.

    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.

    #include <cstdio>
    #include <cmath>
    #include <cstring>
    #include <ctime>
    #include <iostream>
    #include <algorithm>
    #include <set>
    #include <vector>
    #include <sstream>
    #include <queue>
    #include <typeinfo>
    #include <fstream>
    typedef long long ll;
    using namespace std;
    //freopen("D.in","r",stdin);
    //freopen("D.out","w",stdout);
    #define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
    #define maxn 100001
    const int inf=0x7fffffff;   //无限大
    struct node
    {
        int x;
        int y;
    };
    node a[101];
    bool cmp(node k,node m)
    {
        return k.x<m.x;
    }
    int main()
    {
        int n,m;
        cin>>n>>m;
        for(int i=0;i<n;i++)
        {
            cin>>a[i].x;
            a[i].y=i+1;
        }
        sort(a,a+n,cmp);
        queue<int> q;
        int ans=0;
        for(int i=0;i<n;i++)
        {
            if(m>=a[i].x)
            {
                m-=a[i].x;
                q.push(a[i].y);
                ans++;
            }
            else
                break;
        }
        cout<<ans<<endl;
        int first=0;
        while(!q.empty())
        {
            if(first==0)
            {
                cout<<q.front();
                first=1;
                q.pop();
            }
            else
            {
                cout<<" "<<q.front();
                q.pop();
            }
        }
        cout<<endl;
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/qscqesze/p/4245733.html
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