time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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.
Example
inputCopy
3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2
outputCopy
101011
11111
1001
Note
The first test case is described in the problem statement.
In the second test case all numbers from 1 to 5 are beautiful:
if l=3 and r=3 we will have a permutation [1] for m=1;
if l=3 and r=4 we will have a permutation [1,2] for m=2;
if l=2 and r=4 we will have a permutation [3,1,2] for m=3;
if l=2 and r=5 we will have a permutation [3,1,2,4] for m=4;
if l=1 and r=5 we will have a permutation [5,3,1,2,4] for m=5.
#include<iostream>
using namespace std;
const int maxn=2e5+50;
int t,n,x,l,r;
int a[maxn];
int main() {
cin>>t;
while(t--) {
cin>>n;
for(int i=1; i<=n; i++) {
cin>>x;
a[x]=i;
}
l=r=a[1];
for(int i=1; i<=n; i++) {
l=min(l,a[i]);
r=max(r,a[i]);
if(r-l+1==i)
cout<<1;
else
cout<<0;
}
cout<<endl;
}
return 0;
}
