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题目来源：

Description

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.

Sample Input

3

6 4 5 1 3 2 6 5 5 3 1 2 4 4 1 4 3 2

Sample Output

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.

解题思路：

排序后维护最大值位置mx和最小值位置mi，i=0依次遍历，当mx-mi==i时代表成立，反之为0；

具体看代码。

AC代码1：

#include 
   
    using namespace std;#define SIS std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)#define endl '\n'const int MAXN = 2e5+5;pair
    
      arr[MAXN]; int main(){
       SIS;    int T,n,x;    cin >> T;    while(T--)    {
           cin >> n;        int mi=0x3f3f3f3f,mx=0;        for(int i=0;i
     
      > arr[i].first;            arr[i].second=i;        }        sort(arr,arr+n);        vector
      
        v(n);        for(int i=0;i
      
     
    
   

仔细一想本题输入的是1-n的n个不重复的数，所以根本不需要排序。

AC代码2：

#include 
   
    using namespace std;#define SIS std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)#define endl '\n'const int MAXN = 2e5+5;int arr[MAXN];int main(){
       SIS;    int T,n,x;    cin >> T;    while(T--)    {
           cin >> n;        int mi=0x3f3f3f3f,mx=0;        for(int i=1;i<=n;i++)        {
               cin >> x;            arr[x]=i;        }        vector
    
      v(n);        for(int i=1;i<=n;i++)        {
               mi=min(arr[i],mi);            mx=max(arr[i],mx);            if(mx-mi==i-1) v[i-1]=1;            else v[i-1]=0;        }        for(auto x:v) cout << x;        cout << endl;    }    return 0;}
    
   

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