Codeforces Round #346 (Div. 2) C Tanya and Toys

C. Tanya and Toys

题目链接http://codeforces.com/contest/659/problem/C

Description

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, ..., an 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, ..., an (1 ≤ ai ≤ 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, ..., tk (1 ≤ ti ≤ 109) — the types of toys that Tanya should choose.

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

Sample Input

3 7

1 3 4

Sample Output

2

2 5

题意：

给你n个数字，求1～1e9之间除了那n个数字最多可以选多少个不同的数字，使和小于等于m。

题解：

从1开始枚举，只要可以满足条件则从小到大枚举每个可能，贪心即可必然小的必须选。这里使用的bitset速度很快。

代码：

#include<bits/stdc++.h>
using namespace std;
bitset <1000000100> s;
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    int t;
    for (int i = 1;i <= n;i++){scanf("%d",&t);s[t]=1;}
    int c = 0;
    int mm = m;
    for (int i = 1;i <= 1000000000;i++){
        if (s[i])
            continue;
        if (mm < i)
            break;
        mm-=i;
        c++;
    }
    mm = m;
    printf("%d ",c);
    for (int i = 1;i <= 1000000000;i++){
        if (s[i])
            continue;
        if (mm < i)
            break;
        mm-=i;
        printf("%d ",i);
    }
}