Codeforces Round #532 (Div. 2)- B（思维）

最新推荐文章于 2019-03-27 17:07:26 发布

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最新推荐文章于 2019-03-27 17:07:26 发布

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原文链接：http://www.cnblogs.com/Staceyacm/p/10781851.html

本文探讨了Arkady在创建竞赛编程题目时面临的挑战，详细解释了如何通过维护问题池并跟踪不同难度级别的题目来确定何时可以举行一轮比赛。文章提供了一个具体的算法实现，展示了如何在每次创建新题目后判断是否可以立即进行一轮比赛。

Arkady coordinates rounds on some not really famous competitive programming platform. Each round features nn problems of distinct difficulty, the difficulties are numbered from 11 to nn.

To hold a round Arkady needs nn new (not used previously) problems, one for each difficulty. As for now, Arkady creates all the problems himself, but unfortunately, he can't just create a problem of a desired difficulty. Instead, when he creates a problem, he evaluates its difficulty from 11 to nn and puts it into the problems pool.

At each moment when Arkady can choose a set of nn new problems of distinct difficulties from the pool, he holds a round with these problems and removes them from the pool. Arkady always creates one problem at a time, so if he can hold a round after creating a problem, he immediately does it.

You are given a sequence of problems' difficulties in the order Arkady created them. For each problem, determine whether Arkady held the round right after creating this problem, or not. Initially the problems pool is empty.

Input

The first line contains two integers nn and mm (1≤n,m≤1051≤n,m≤105) — the number of difficulty levels and the number of problems Arkady created.

The second line contains mm integers a1,a2,…,ama1,a2,…,am (1≤ai≤n1≤ai≤n) — the problems' difficulties in the order Arkady created them.

Output

Print a line containing mm digits. The ii-th digit should be 11 if Arkady held the round after creation of the ii-th problem, and 00 otherwise.

Examples

input

Copy

3 11
2 3 1 2 2 2 3 2 2 3 1

output

Copy

00100000001

input

Copy

4 8
4 1 3 3 2 3 3 3

output

Copy

00001000

Note

In the first example Arkady held the round after the first three problems, because they are of distinct difficulties, and then only after the last problem.

思路：一开始理解错题意了，以前剩下的题目也要算在后边，故可以用标记数组的量来检测

代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>

using namespace std;
int vis[100005],a[100005];
int main()
{
	
	int n,m;
	cin>>n>>m;
	int sum=0;
	for(int t=0;t<m;t++)
	{
		scanf("%d",&a[t]);
		if(vis[a[t]]==0)
		{
			sum++;
			
		}
		vis[a[t]]++;
		
		
		if(sum==n)
		{
			printf("1");
			sum=0;
			for(int j=1;j<=n;j++)
			{
				vis[j]--;
				if(vis[j]>0)
				{
				sum++;	
			
				}
			}
			
		}
		else if(sum<n)
		{
			printf("0");
		}
		
		
	}
	return 0;
}

转载于:https://www.cnblogs.com/Staceyacm/p/10781851.html