CF1659D.Reverse Sort Sum

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题目描述

Suppose you had an array AA of nn elements, each of which is 00 or 11 .

Let us define a function f(k,A)f(k,A) which returns another array BB , the result of sorting the first kk elements of AA in non-decreasing order. For example, f(4,[0,1,1,0,0,1,0])=[0,0,1,1,0,1,0]f(4,[0,1,1,0,0,1,0]) = [0,0,1,1,0,1,0] . Note that the first 44 elements were sorted.

Now consider the arrays B1,B2,,BnB_1, B_2,\ldots, B_n generated by f(1,A),f(2,A),,f(n,A)f(1,A), f(2,A),\ldots,f(n,A) . Let CC be the array obtained by taking the element-wise sum of B1,B2,,BnB_1, B_2,\ldots, B_n .

For example, let A=[0,1,0,1]A=[0,1,0,1] . Then we have B1=[0,1,0,1]B_1=[0,1,0,1] , B2=[0,1,0,1]B_2=[0,1,0,1] , B3=[0,0,1,1]B_3=[0,0,1,1] , B4=[0,0,1,1]B_4=[0,0,1,1] . Then C=B1+B2+B3+B4=[0,1,0,1]+[0,1,0,1]+[0,0,1,1]+[0,0,1,1]=[0,2,2,4]C=B_1+B_2+B_3+B_4=[0,1,0,1]+[0,1,0,1]+[0,0,1,1]+[0,0,1,1]=[0,2,2,4] .

You are given CC . Determine a binary array AA that would give CC when processed as above. It is guaranteed that an array AA exists for given CC in the input.

输入格式

The first line contains a single integer tt ( 1t10001 \leq t \leq 1000 ) — the number of test cases.

Each test case has two lines. The first line contains a single integer nn ( 1n21051 \leq n \leq 2 \cdot 10^5 ).

The second line contains nn integers c1,c2,,cnc_1, c_2, \ldots, c_n ( 0cin0 \leq c_i \leq n ). It is guaranteed that a valid array AA exists for the given CC .

The sum of nn over all test cases does not exceed 21052 \cdot 10^5 .

输出格式

For each test case, output a single line containing nn integers a1,a2,,ana_1, a_2, \ldots, a_n ( aia_i is 00 or 11 ). If there are multiple answers, you may output any of them.

输入输出样例

  • 输入#1

    5
    4
    2 4 2 4
    7
    0 3 4 2 3 2 7
    3
    0 0 0
    4
    0 0 0 4
    3
    1 2 3

    输出#1

    1 1 0 1 
    0 1 1 0 0 0 1 
    0 0 0 
    0 0 0 1 
    1 0 1

说明/提示

Here's the explanation for the first test case. Given that A=[1,1,0,1]A=[1,1,0,1] , we can construct each BiB_i :

  • B1=[1,1,0,1]B_1=[\color{blue}{1},1,0,1] ;
  • B2=[1,1,0,1]B_2=[\color{blue}{1},\color{blue}{1},0,1] ;
  • B3=[0,1,1,1]B_3=[\color{blue}{0},\color{blue}{1},\color{blue}{1},1] ;
  • B4=[0,1,1,1]B_4=[\color{blue}{0},\color{blue}{1},\color{blue}{1},\color{blue}{1}]

And then, we can sum up each column above to get C=[1+1+0+0,1+1+1+1,0+0+1+1,1+1+1+1]=[2,4,2,4]C=[1+1+0+0,1+1+1+1,0+0+1+1,1+1+1+1]=[2,4,2,4] .

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