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

C. Tanya and Toys

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

In Berland recently a new collection of toys went on sale. This collection consists of 109 types
of toys, numbered with integers from 1 to109.
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.

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.

题意：给出 n 个 Tanya 已经拥有玩具类型编号和 m 个单位的钱，第 i 个类型的玩具需要 i 个单位的钱，

你需要在给定的上限为 m 个单位的钱的条件下尽可能多地挑 Tanya 未拥有类型的玩具进行购买，

买的玩具类型数要求最多，输出 类型总数 及 各个购买的玩具类型编号。

思路：直接从编号小到大遍历，购买未拥有的玩具类型。

易错点：因为数据很大，数组开不了，所以后来没用下标数组，

直接存储已经购买的玩具类型编号来处理，不会 RE 了。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
#define size 120000
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;
ll a[size], ans[size];
int main()
{
#ifdef OFFLINE
freopen("t.txt", "r", stdin);
#endif
ll n, m, i, j, type, sum, k, p;
while (~scanf("%lld%lld", &n, &m))
{
memset(a, -1, sizeof(a));
k = 1;
for (i = 1; i <= n; ++i)
{
scanf("%lld", &a[k]);//直接存储已经拥有的玩具类型号
k++;
}
sort(a + 1, a + k);//小到大排序玩具类型号便于下面比较
type = 0;
i = p = 1;
while (m && m >= i)//钱数允许
{
if (i != a[p])//购买还未拥有的玩具类型
{
type++;
m -= i;
ans[type] = i;
}
else
p++;//下一个已经拥有的玩具类型号在数组的下标
i++;//前往后扫，保证最大程度购买不同类型玩具
}
printf("%lld\n", type);
for (i = 1; i <= type; ++i)
{
if (i == type)
printf("%lld\n", ans[i]);
else
printf("%lld ", ans[i]);
}
}
}