Codeforces 659C Tanya and Toys【贪心】

最新推荐文章于 2018-07-23 16:39:25 发布

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最新推荐文章于 2018-07-23 16:39:25 发布

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分类专栏：贪心文章标签： Codeforces 659C

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本文介绍了一个玩具选购的问题背景及解决思路，旨在帮助理解如何在有限预算内购买最多数量且不重复的玩具，通过排序和逐个检查的方式实现了最优解。

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In Berland recently a new collection of toys went on sale. This collection consists of 10⁹ types of toys, numbered with integers from 1 to 10⁹. A toy from the new collection of the i-th type costs i bourles.

Tania has managed to collect n different types of toys a₁, a₂, ..., a_n from the new collection. Today is Tanya's birthday, and her mother decided to spend no more than m bourles on the gift to the daughter. Tanya will choose several different types of toys from the new collection as a gift. Of course, she does not want to get a type of toy which she already has.

Tanya wants to have as many distinct types of toys in her collection as possible as the result. The new collection is too diverse, and Tanya is too little, so she asks you to help her in this.

Input

The first line contains two integers n (1 ≤ n ≤ 100 000) and m (1 ≤ m ≤ 10⁹) — the number of types of toys that Tanya already has and the number of bourles that her mom is willing to spend on buying new toys.

The next line contains n distinct integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹) — the types of toys that Tanya already has.

Output

In the first line print a single integer k — the number of different types of toys that Tanya should choose so that the number of different types of toys in her collection is maximum possible. Of course, the total cost of the selected toys should not exceed m.

In the second line print k distinct space-separated integers t₁, t₂, ..., t_k (1 ≤ t_i ≤ 10⁹) — the types of toys that Tanya should choose.

If there are multiple answers, you may print any of them. Values of t_i can be printed in any order.

Examples

Input

3 7
1 3 4

Output

2
2 5

Input

4 14
4 6 12 8

Output

4
7 2 3 1

Note

In the first sample mom should buy two toys: one toy of the 2-nd type and one toy of the 5-th type. At any other purchase for 7 bourles (assuming that the toys of types 1, 3 and 4 have already been bought), it is impossible to buy two and more toys.

题目大意：

有编号从1-10^9的玩具，想要买一个玩具就要花费对应编号的价钱，现在我们初始的时候已经有了N个玩具和有的钱数M，问如何购买能够购买最多的玩具，输出任意解即可。

思路;

1、将已经有的玩具的编号进行从小到大排序。然后判断，我们初始的金钱数为M，M最大为10^9，简单判断一下，如果我们有最大金额数10^9，那么如果我们从编号为1的玩具开始购买，简单分析可知，10^5以后的编号的玩具就是一定不能购买的到的了(（1+10^5）*10^5）/2>10^9.

2、那么我们从编号为1的玩具开始判断，一直判断到10^5的编号的玩具为止，如果当前玩具已经有了，那么跳过，否则购买即可。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[100050];
int ans[100050];
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
        sort(a,a+n);
        int now=0;
        int cont=0;
        for(int i=1;i<=1000000;i++)
        {
            if(a[now]!=i||now>=n)
            {
                if(m>=i)
                {
                    m-=i;
                    ans[cont++]=i;
                }
                else break;
            }
            if(a[now]==i)
            {
                while(1)
                {
                    if(a[now]==i)now++;
                    if(now>=n)break;
                    else break;
                }
            }
        }
        printf("%d\n",cont);
        for(int i=0;i<cont;i++)
        {
            printf("%d ",ans[i]);
        }
        printf("\n");
    }
}