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

本文介绍了一个玩具选购的问题背景,包括玩具的种类与价格设定,并提出了在有限预算内如何为已拥有部分玩具的孩子选购新玩具以最大化玩具种类的方法。通过调整循环结构避免了超时问题。

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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 13 and 4 have already been bought), it is impossible to buy two and more toys.

题目链接

题意:给你一个n表示已有的n种类型的玩具第二行n个数表示玩具的类型  求价值小于等于m能找到的玩具的最多种类 

题意很简单  但是10^9数据很大  用数组标记肯定超内存 而用两层for循环会超时    所以这里把for循环位置调换  就不会超时了

代码:

#include<stdio.h>
#include<algorithm>
int a[1000001]={0};
int s[1000001]={0};
using namespace std;
int main()
{
    int n,m,sum=0;
    scanf("%d%d",&n,&m);
    for(int i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
    }
    sort(a,a+n);
    a[n]=1000000000;
//    for(int i=0;i<=n;i++)
//      printf("%d ",a[i]);
    int t=1,k=0;
     int flag=0;
    for(int i=0;i<=n;i++)
    {
        for(int j=t;j<=1000000000;j++)
        {
            if(j!=a[i])
            {
                sum+=j;
                if(sum>m)
                {
                    flag=1;
                    break;
                }
                s[k++]=j;
            }
            else
            {
                break;
            }

        }
        if(flag==1)
        {
           break;
        }
         t=a[i]+1;
    }
    printf("%d\n",k);
    for(int i=0;i<k;i++)
    {
        printf("%d ",s[i]);
    }
    printf("\n");
}


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