B. Build a Contest
time limit per test:1 second
memory limit per test:256 megabytes
inputstandard input
outputstandard output
Arkady coordinates rounds on some not really famous competitive programming platform. Each round features nn problems of distinct difficulty, the difficulties are numbered from 1 to n.
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 1 to n 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 n and m (1≤n,m≤10^5) — the number of difficulty levels and the number of problems Arkady created.
The second line contains m integers a1,a2,…,am (1≤ai≤n) — the problems’ difficulties in the order Arkady created them.
Output
Print a line containing m digits. The i-th digit should be 1 if Arkady held the round after creation of the i-th problem, and 0 otherwise.
Examples
input
3 11
2 3 1 2 2 2 3 2 2 3 1
output
00100000001
input
4 8
4 1 3 3 2 3 3 3
output
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：从前到后搜索，每n个不同的难度级别输出1，否则为0；
注意：4 1 2 4 1 2 3 3的输出应该是00000011。
#include<iostream>
#include<stdlib.h>
#include<cstring>
using namespace std;
int main()
{
long long int n,m,a[100005],b[100005]={0},t,i,cnt=0,j;
cin>>n>>m;
for(i=0;i<m;i++)
{
cin>>a[i];
t=a[i];
if(b[t]==0)
{
b[t]++;
cnt++;
if(cnt==n)
{
cout<<"1";
for(j=1;j<=n;j++)
{
b[j]--;
if(b[j]==0)
cnt--;
}
}
else
{
cout<<"0";
}
}
else
{
b[t]++;
cout<<"0";
}
}
cout<<endl;
return 0;
}