B. Build a Contest
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard 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 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.
看题很头疼，不知道怎么看，心累，开始以为是只要满足不一样的n个数就吧前面的数字全部删掉，结果是前面的数字每个都删掉一个。无奈不会写pair。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int n,m;
int a[100005],cnt[100005],ans[100005];
int main(){
scanf("%d%d",&n,&m);
int num=0;
memset(ans,0,sizeof(ans));
for(int i=0;i<m;i++){
scanf("%d",&a[i]);
if(cnt[a[i]]==0) num++;
cnt[a[i]]++;
if(num==n){
ans[i]=1;
num=0;
for(int j=1;j<=n;j++){
cnt[j]--;
if(cnt[j]>0) num++;
}
}
}
for(int i=0;i<m;i++){
printf("%d",ans[i]);
}
printf("\n");
return 0;
}#include<cstdio>
#include<set>
using namespace std;
int a[100005],num[100005];
int main() {
int n,m,t=0;
set<pair<int,int> > s;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++) s.insert(pair<int,int>{0,i});
for(int i=1;i<=m;i++) {
scanf("%d",&a[i]);
s.erase(pair<int,int>{num[a[i]],a[i]});
s.insert(pair<int,int>{++num[a[i]],a[i]});
set<pair<int,int> >::iterator it=s.begin();
if(it->first>t) printf("1"),t++;
else printf("0");
}
}