Codeforces Round #346 (Div. 2) C. Tanya and Toys
来源:互联网 发布:教师课件制作的软件 编辑:程序博客网 时间:2024/05/24 23:13
题目链接:点击打开链接
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.
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.
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.
3 71 3 4
22 5
4 144 6 12 8
47 2 3 1
思路:肯定是尽可能的拿编号小的,同时已经有的不能拿
代码:
#include<stdio.h>#include<algorithm>using namespace std;int main(){ int n,m,a[100010],i,j,b[100010]; scanf("%d%d",&n,&m); for(i=1;i<=n;i++) scanf("%d",&a[i]); sort(a+1,a+1+n); j=1;int len=0,sum=0; for(i=1;1;i++) { if(sum>m) break; if(j==n+1||a[j]!=i) { b[len++]=i; sum+=i; } else if(a[j]==i) { j++; } } len--; printf("%d\n",len); for(i=0;i<len;i++) printf("%d ",b[i]); printf("\n");}
- Codeforces Round #346 (Div. 2) C. Tanya and Toys
- Codeforces Round #346 (Div. 2)--C. Tanya and Toys
- Codeforces Round #346 (Div. 2) C.Tanya and Toys
- Codeforces Round #346 (Div. 2) C. Tanya and Toys
- Codeforces Round #346 (Div. 2) - C Tanya and Toys
- Codeforces Round #346 (Div. 2)C. Tanya and Toys
- Codeforces Round #346 (Div. 2) C. Tanya and Toys
- Codeforces Round #346 (Div. 2) C. Tanya and Toys 贪心
- Codeforces Round #346 (Div. 2)-C. Tanya and Toys(贪心)
- 【CodeForces】[659C]Tanya and Toys
- Codeforces 659C Tanya and Toys 【水题】
- CodeForces-659C-Tanya and Toys
- CodeForces 659 C. Tanya and Toys
- Codeforces 659C Tanya and Toys
- Codeforces 659C Tanya and Toys【贪心】
- Codeforces 659C Tanya and Toys 【贪心】
- C. Tanya and Toys
- C - Tanya and Toys
- java se 集合 笔记
- 实现一个线程池
- 怎样修改编辑PDF文件
- jQuery仪表盘指示器动画插件 6种仪表样式
- jQuery EasyUI使用教程之在数据网格的页脚中显示摘要信息
- Codeforces Round #346 (Div. 2) C. Tanya and Toys
- ionic集成极光推送插件-iOS
- Swift 单例
- zookeeper分布式协调服务的使用一
- 好险,差点跳进自已挖的坑
- 有用的APP开发
- 关于vs2015不能使用inet_addr等函数
- JavaSE----API之常用类(Object、Scanner、String)
- 问题集合 ---- linux 静态库和动态库创建、检查、使用建议