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