Codeforces Round #654 (Div. 2)——D. Grid-00100

## D. Grid-00100
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®−min®)²+(max©−min©)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.
Input
The input consists of multiple test cases. The first line contains a single integer tt (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.
Output
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.
Example
input

4
2 2
3 8
1 0
4 16

output

0
10
01
2
111
111
101
0
0
0
1111
1111
1111
1111

题意思路：给一个n，一个k，n表示n行n列的矩阵，k表示矩阵中1的个数，（矩阵中只存在0或者1）。然后求出最小的行和最大值-行和最小值的结果的平方与最小的列和最大值-列和最小值的结果的平方的加和。然后输出该矩阵。

f(A)=(max®−min®)²+(max©−min©)²

思路的话是要行或者列的最大值和最小值得差值最小，那么则需要平均的分给每一行每一列，，注意是同时平均分给每一行每一列，所以就斜着平均的分配（从左上角到右下角）。然后根据分析可得，行列和的差值只有两种，一种是每行或每列得职全都一样（0），另一种是最大和最小差值为1，行列加在一起就是（2）。

代码

#include 
using namespace std;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,k;
        scanf("%d%d",&n,&k);
        int tmp=k%n==0?0:2;
        printf("%d\n",tmp);
        vector<vector<int>>a(n,vector<int>(n));
        for (int i=0; i<n; ++i)
            for (int j=0; j<n; ++j)
                if (k>0)
                    a[j][(i+j)%n]=1,--k;
        for (int i=0; i<n; ++i)
        {
            for (int j=0; j<n; ++j)
                printf("%d",a[i][j]);
            printf("\n");
        }
    }
    return 0;
}