B. Build a Contest
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.
题意: 有 n个问题 , Arkady 创造 m 道问题 , 每一轮当完成 n 道 题的时候输出 1 ,否则输出 0 . 然后进行下一轮,但是Arkady 每一轮难度为 n 的问题只能完成一个 .
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <stack>
#include <queue>
#define Swap(a,b) a ^= b ^= a ^= b
using namespace std ;
const int MAX = 100005;
const int inf = 0xffffff;
typedef long long LL;
int a[MAX] ;
int book[MAX];
int main()
{
int n ,m ;
int ans = 0 ;
cin >> n >> m ;
for(int i = 1 ; i<=m ; i++ )
{
cin >> a[i] ;
}
int cnt = 0 ;
for(int i = 1 ; i<=m ;i++)
{
if(!book[a[i]])
{
book[a[i]]++ ;
cnt++ ;
if(cnt == n )
{
cout<<"1";
// 再把这些问题(n 个 )从问题池中除去 ;
// 剩下的下一轮中还会用 ;
for(int j = 1 ; j<=n ; j++)
{
book[j]-- ;
if(book[j]==0)
cnt-- ;
}
}
else
{
cout<<"0";
}
}
else
{
book[a[i]]++ ;
cout<<"0";
}
}
return 0 ;
}
