Codeforces Round #604 (Div. 2) B题

Beautiful Numbers

You are given a permutation p=[p1,p2,…,pn] of integers from 1 to n. Let’s call the number m (1≤m≤n) beautiful, if there exists two indices l,r (1≤l≤r≤n), such that the numbers [pl,pl+1,…,pr] is a permutation of numbers 1,2,…,m.

For example, let p=[4,5,1,3,2,6]. In this case, the numbers 1,3,5,6 are beautiful and 2,4 are not. It is because:

if l=3 and r=3 we will have a permutation [1] for m=1;
if l=3 and r=5 we will have a permutation [1,3,2] for m=3;
if l=1 and r=5 we will have a permutation [4,5,1,3,2] for m=5;
if l=1 and r=6 we will have a permutation [4,5,1,3,2,6] for m=6;
it is impossible to take some l and r, such that [pl,pl+1,…,pr] is a permutation of numbers 1,2,…,m for m=2 and for m=4.
You are given a permutation p=[p1,p2,…,pn]. For all m (1≤m≤n) determine if it is a beautiful number or not.

Input
The first line contains the only integer t (1≤t≤1000) — the number of test cases in the input. The next lines contain the description of test cases.

The first line of a test case contains a number n (1≤n≤2⋅105) — the length of the given permutation p. The next line contains n integers p1,p2,…,pn (1≤pi≤n, all pi are different) — the given permutation p.

It is guaranteed, that the sum of n from all test cases in the input doesn’t exceed 2⋅105.

Output
Print t lines — the answers to test cases in the order they are given in the input.

The answer to a test case is the string of length n, there the i-th character is equal to 1 if i is a beautiful number and is equal to 0 if i is not a beautiful number.

---

思维题，输入时我们记录每个数字的位置，然后枚举 1–m ，如果这些数的位置是不间断的，那么 m 就可以取得到；

代码：

#include
#define LL long long
#define pa pair
#define lson k<<1
#define rson k<<1|1
#define inf 0x3f3f3f3f
//ios::sync_with_stdio(false);
using namespace std;
const int N=100100;
const int M=4001000;
const LL mod=998244353;
int p[200100];
int main(){
	ios::sync_with_stdio(false);
	int t;
	cin>>t;
	while(t--){
		int n;
		cin>>n;
		int x;
		for(int i=1;i<=n;i++){
			cin>>x;
			p[x]=i;
		}
		int mmin=p[1],mmax=p[1];
		for(int i=1;i<=n;i++){
			mmin=min(p[i],mmin);
			mmax=max(p[i],mmax);
			if(mmax-mmin+1==i) cout<<1;
			else cout<<0;
		}
		cout<;
	}
	return 0;
}