A. Doremy‘s IQ-- Codeforces Round ＃808 (Div. 1)

Problem - 1707A - Codeforces

A. Doremy's IQ

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Doremy is asked to test nn contests. Contest ii can only be tested on day ii. The difficulty of contest ii is aiai. Initially, Doremy's IQ is qq. On day ii Doremy will choose whether to test contest ii or not. She can only test a contest if her current IQ is strictly greater than 00.

If Doremy chooses to test contest ii on day ii, the following happens:

• if ai>qai>q, Doremy will feel she is not wise enough, so qq decreases by 11;
• otherwise, nothing changes.

If she chooses not to test a contest, nothing changes.

Doremy wants to test as many contests as possible. Please give Doremy a solution.

Input

The input consists of multiple test cases. The first line contains a single integer tt (1≤t≤1041≤t≤104) — the number of test cases. The description of the test cases follows.

The first line contains two integers nn and qq (1≤n≤1051≤n≤105, 1≤q≤1091≤q≤109) — the number of contests and Doremy's IQ in the beginning.

The second line contains nn integers a1,a2,⋯,ana1,a2,⋯,an (1≤ai≤1091≤ai≤109) — the difficulty of each contest.

It is guaranteed that the sum of nn over all test cases does not exceed 105105.

Output

For each test case, you need to output a binary string ss, where si=1si=1 if Doremy should choose to test contest ii, and si=0si=0 otherwise. The number of ones in the string should be maximum possible, and she should never test a contest when her IQ is zero or less.

If there are multiple solutions, you may output any.

Example

input

Copy

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

output

Copy

1
11
110
1110
01111

Note

In the first test case, Doremy tests the only contest. Her IQ doesn't decrease.

In the second test case, Doremy tests both contests. Her IQ decreases by 11 after testing contest 22.

In the third test case, Doremy tests contest 11 and 22. Her IQ decreases to 00 after testing contest 22, so she can't test contest 33.

=========================================================================

贪心

正确性证明   设   1..2.....pos  .....................n  设我们在pos,位置遇到一个大于当前p的，如果我们执行了减1操作，今后再也遇不见大于p的话，没有什么影响。然而一旦遇见一个pos1大于当前p

 1..2.....pos  ........pos1.............n 

pos1+1----n之间，无论我们执行pos,还是pos1,所得结果都是一样的，因为p都变成了p-1

而pos-pos1之间，如果我们先执行了pos,操作，那么显然不如执行pos1操作更优

所以我们要尽可能的吧-1操作留在末尾，取p能够满足最长的末尾段，全部涂上1，之前的在p之内的再涂上1.

用二分比较容易理解，网上普遍做法是二分的逆，也就是从最末尾往前推连续一段，设p最初为0，遇见比p大的就加1，一直加到p=p0即可。其实这里的p没有任何实际意义，更多的是一种数学表示，是抽象出来的一个数。按照这样我们推导p=p0时，p就完全覆盖了后面的这一段。很多题解说p就是当前p,其实并不准确

例如

p0=5

序列为  5 4 3 2 1   p从0一直变到5，并不代表我们p0的变化，即p仅仅是一个没有意义抽象出来的概念。

#include <bits/stdc++.h>
using namespace std;

int a[100000+10];
bool book[100000+10];

int  main()
{


   int t;

   cin>>t;

   while(t--)
   {
       int n,q;

       cin>>n>>q;

       for(int i=1;i<=n;i++)
       {
           cin>>a[i];
           book[i]=0;
       }

       int num=0;

       for(int i=n;i>=1;i--)
       {
           if(num<a[i])
           {
               if(num+1<=q)
                {
                    num++;
                    book[i]=1;
                }

           }
           else
           {

               book[i]=1;

           }
       }

       for(int i=1;i<=n;i++)
       {
           cout<<book[i];
       }
       cout<<endl;

   }
    return 0;
}