CodeForces-659C-Tanya and Toys
2016-04-06 20:42
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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
Input
Output
Input
Output
有n种已经拥有的物品和m元钱,第i种物品价格为i。问你最多可以拥有的物品种类。其实也就是问没有选过的toy能最多选多少个。
很简单的贪心,从最小的找起就好了,但是1e9有点大,数组开不下,用了vector
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
Input
3 7 1 3 4
Output
2 2 5
Input
4 14 4 6 12 8
Output
4 7 2 3 1
有n种已经拥有的物品和m元钱,第i种物品价格为i。问你最多可以拥有的物品种类。其实也就是问没有选过的toy能最多选多少个。
很简单的贪心,从最小的找起就好了,但是1e9有点大,数组开不下,用了vector
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<vector> using namespace std; vector <int> ans; bool vis[1111111]; int main() { int n,m,num; while(scanf("%d%d",&n,&m)!=EOF) { memset(vis,0,sizeof(vis)); ans.clear(); for(int i=0;i<n;i++) { scanf("%d",&num); if (num <= 1000000) vis[num]=1; } for(int i=1;m-i>=0;i++) { if(!vis[i]) { m-=i; ans.push_back (i); } } printf ("%d\n", ans.size ()); for (int i = 0; i < ans.size (); i++) { printf ("%d%c", ans[i], i == (ans.size ()-1) ? '\n':' '); } } return 0; }
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