CF1371D.Grid-00100
普及/提高-
通过率:0%
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题目描述
A mad scientist Dr.Jubal has made a competitive programming task. Try to solve it!
You are given integers n,k . Construct a grid A with size n×n consisting of integers 0 and 1 . The very important condition should be satisfied: the sum of all elements in the grid is exactly k . In other words, the number of 1 in the grid is equal to k .
Let's define:
- Ai,j as the integer in the i -th row and the j -th column.
- Ri=Ai,1+Ai,2+...+Ai,n (for all 1≤i≤n ).
- Cj=A1,j+A2,j+...+An,j (for all 1≤j≤n ).
- In other words, Ri are row sums and Cj are column sums of the grid A .
- For the grid A let's define the value f(A)=(max(R)−min(R))2+(max(C)−min(C))2 (here for an integer sequence X we define max(X) as the maximum value in X and min(X) as the minimum value in X ).
Find any grid A , which satisfies the following condition. Among such grids find any, for which the value f(A) is the minimum possible. Among such tables, you can find any.
输入格式
The input consists of multiple test cases. The first line contains a single integer t ( 1≤t≤100 ) — the number of test cases. Next t lines contain descriptions of test cases.
For each test case the only line contains two integers n , k (1≤n≤300,0≤k≤n2) .
It is guaranteed that the sum of n2 for all test cases does not exceed 105 .
输出格式
For each test case, firstly print the minimum possible value of f(A) among all tables, for which the condition is satisfied.
After that, print n lines contain n characters each. The j -th character in the i -th line should be equal to Ai,j .
If there are multiple answers you can print any.
输入输出样例
输入#1
4 2 2 3 8 1 0 4 16
输出#1
0 10 01 2 111 111 101 0 0 0 1111 1111 1111 1111
说明/提示
In the first test case, the sum of all elements in the grid is equal to 2 , so the condition is satisfied. R1=1,R2=1 and C1=1,C2=1 . Then, f(A)=(1−1)2+(1−1)2=0 , which is the minimum possible value of f(A) .
In the second test case, the sum of all elements in the grid is equal to 8 , so the condition is satisfied. R1=3,R2=3,R3=2 and C1=3,C2=2,C3=3 . Then, f(A)=(3−2)2+(3−2)2=2 . It can be proven, that it is the minimum possible value of f(A) .