CSP201503-1: image rotation, csp201503-1 Rotation

CSP201503-1: image rotation, csp201503-1 Rotation

Introduction:CSP (http://www.cspro.org/lead/application/ccf/login.jsp) is a "Computer vocational qualification certification" examination initiated by CCF, perform capability certification for professionals in computer software development, software testing, information management, and other fields. The subjects of certification are those who are engaged in or will be engaged in IT professional technical and technical management personnel, as well as review subjects for university recruitment and postgraduate students.

 

• Problem description

Rotation is the basic operation of image processing. In this case, you need to rotate an image 90 degrees counter-clockwise.

The image representation in the computer can be expressed by a matrix. to rotate an image, you only need to rotate the corresponding matrix.

• Input Format

The first line contains two integers, n and m, indicating the number of rows and columns of the image matrix.

Each line in the next n rows contains m integers, indicating the input image.

• Output Format

Output m rows. Each row contains n integers, indicating the matrix after 90 degrees of counter-clockwise rotation of the original matrix.

• Sample Input

  2 3

  1 5 3

  3 2 4

• Sample output

  3 4

  5 2

  1 3

• Scale and conventions of evaluation cases

  1 ≤ n, m ≤ 1,000, and the number in the matrix is a non-negative integer not greater than 1000.

 

• Source code (1): Use a dynamically allocated one-dimensional array

/* One-dimensional array solution, dynamic space allocation */ # Include <stdio. h> # Include <stdlib. h> # Include <memory. h>   Int main (void) { Int n; // number of rows Int m; // Number of Columns Scanf ("% d", & n ); Scanf ("% d", & m ); Int count = m * n + 1; Int * input = (int *) malloc (sizeof (int) * count ); Memset (input, 0, sizeof (int) * count ); Int * output = (int *) malloc (sizeof (int) * count ); Memset (output, 0, sizeof (int) * count ); For (int I = 1; I <count; I ++) { Scanf ("% d", input + I ); Int row_in = (I/m) + 1; Int column_in = I % m; If (column_in = 0) { Row_in-= 1; Column_in = m; } Int row_out = m + 1-column_in; Int column_out = row_in; Output [(row_out-1) * n + column_out] = input [I]; }   Int num = 0; For (int I = 1; I <count; I ++) { Num + = 1; If (num = n) { Printf ("% d \ n", output [I]); Num = 0; } Else { Printf ("% d", output [I]); } } Free (input ); Free (output ); Return 0; }

 

• Source code (2): Use a dynamically allocated two-dimensional array

/* Two-dimensional array solution, dynamic space allocation */ # Include <stdio. h> # Include <stdlib. h> # Include <memory. h>   Int main (void) { Int n; // number of rows Int m; // Number of Columns Scanf ("% d", & n ); Scanf ("% d", & m ); // Allocate space for the input two-dimensional data Int ** input = (int **) malloc (sizeof (int *) * n); // allocate space for rows For (int I = 0; I <n; I ++) // allocate space to the column { Input [I] = (int *) malloc (sizeof (int) * m ); Memset (input [I], 0, sizeof (int) * m ); } // Allocate space to the output two-dimensional array Int ** output = (int **) malloc (sizeof (int *) * m); // allocate space for rows For (int I = 0; I <m; I ++) // allocate space to the column { Output [I] = (int *) malloc (sizeof (int) * n ); Memset (output [I], 0, sizeof (int) * n ); } For (int I = 0; I <n; I ++) { For (int j = 0; j <m; j ++) { Scanf ("% d", & input [I] [j]); Output [m-1-j] [I] = input [I] [j]; } }   For (int I = 0; I <m; I ++) { For (int j = 0; j <n; j ++) { If (j = N-1) { Printf ("% d \ n", output [I] [j]); } Else { Printf ("% d", output [I] [j]); } } } // Reclaim the memory allocated to the input Array For (int I = 0; I <n; I ++) { Free (input [I]); } Free (input ); // Reclaim the memory allocated to the output Array For (int I = 0; I <m; I ++) { Free (output [I]); } Free (output ); Return 0; }