02: diagonal grids of the same column, 02 diagonal Grids

02: diagonal grids of the same column, 02 diagonal Grids
02: diagonal grids of the same column

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Total time limit:

1000 ms

 

Memory limit:

65536kB

Description

Input three natural numbers N, I, j (1 <= I <= N, 1 <= j <= N ), output in a N * N lattice Board (the columns and columns start from 1), with the position of all grids with the grid (I, j), the same column, and the same diagonal line.

For example, n = 4, I = 2, j = 3 indicates the lattice of the third column in the second row of the Board, for example:

First column	Column 2	Column 3	Column 4
				First line
		(2, 3)		Row 2
				Row 3
				Row 4

When n = 4, I = 2, j = 3, the output result is:

(2, 1) (2, 2) (2, 3) (2, 4) Position of the grid on the same row

(1, 3) (2, 3) (3, 3) (4, 3) Position of the grid in the same column

(1, 2) (2, 3) (3, 4) Position of the grid on the upper left to the lower right corner

() Position of the grid on the lower left to the upper right corner

 

Input

One row, three natural numbers N, I, j, and two adjacent numbers are separated by a single space. 1 <= N <= 10.

Output

Four rows:
First line: outputs the grid position of the same row from left to right;
Row 2: output the grid position of the same column from top to bottom;
Row 3: outputs the same diagonal grid position from top left to bottom right;
Row 4: output the grid position of the same diagonal line from bottom left to top right.

Each grid position is output in the following format: (x, y), x is the row number, y is the column number, and English Punctuation is used. There is no space in the middle.
Separate two adjacent grids with a single space.

Sample Input

4 2 3

Sample output

(2,1) (2,2) (2,3) (2,4)(1,3) (2,3) (3,3) (4,3)(1,2) (2,3) (3,4)(4,1) (3,2) (2,3) (1,4)

Source

Question 2 of NOIP1996 semi-finals popularity Group

 1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<algorithm> 5 using namespace std; 6 int a[10001][10001]; 7 int b[10001][10001]; 8 int ans[10001][10001]; 9 int now;10 int tot;11 int main()12 {13     int n,x,y;14     cin>>n>>x>>y;15     for(int i=1;i<=n;i++)16     {17         for(int j=1;j<=n;j++)18         {19             if(i==x)20             cout<<"("<<i<<","<<j<<")"<<" ";21         }22     }23     cout<<endl;24     for(int i=1;i<=n;i++)25     {26         for(int j=1;j<=n;j++)27         {28             if(j==y)29             cout<<"("<<i<<","<<j<<")"<<" ";30         }31     }32     cout<<endl;33     for(int i=1;i<=n;i++)34     {35         for(int j=1;j<=n;j++)36         {37             if((j-i)==(y-x))38             cout<<"("<<i<<","<<j<<")"<<" ";39         }40     }41     cout<<endl;42     for(int i=n;i>=1;i--)43     {44         for(int j=n;j>=1;j--)45         {46             if((i+j)==(y+x))47             cout<<"("<<i<<","<<j<<")"<<" ";48         }49     }50     return 0;51 }