Letter games and typing games

Letter games and typing games

[Problem description]

Peter liked playing the letter game, so he wrote an interesting game. The rules of the game are to fill in letters in a (N-1) * N table, rules: For each input N, starting from the upper left corner of the (N-1) * N letter area, enter A letter area counterclockwise from letter A. if the product is greater than 26, it will continue to be filled by A, B, and C in the new group. No border is required.

[Input]

Multiple groups of test data (less than 100 groups of data ). Only one row of test data in each group is an integer N (1 <= N <= 30), indicating the table width.

[Output]

For each group of input data, the output N-1 row is the filled table (N-1 rows, N integers per line, separated by spaces ).

[Example input]

5

[Sample output]

It represents a 4*5 clockwise letter area. Its style is as follows:

A B C D E

N O P Q F

M T S R G

L K J I H

Code:

# Include <stdio. h> # define MAX 100 char r [MAX] [MAX]; // r [] is used for filling in data // recursive call. The void solve (int x, int y, char ch, int m, int n)/* parameter description: x, y indicates the starting coordinate. ch indicates the current letter, m, n is the number of rows and columns to be filled in */{int I; // if m is 0, it indicates that the rectangle has been filled in if (m = 0) during the first recursion) {return;} // when m is set to 1, it indicates that one row and two columns are left unfilled, And if (m = 1) {for (I = 1; I <= n; I ++) {r [x] [y] = ch; y ++; ch ++; if (ch = 'Z' + 1) {ch = 'A' ;}}return;} // enter the row for (I = 1; I <n; I ++) {r [x] [y] = ch; y ++; ch ++; if (ch = 'Z' + 1) {ch = 'A ';}} // fill in the column for (I = 1; I <m; I ++) {r [x] [y] = ch; x ++; ch ++; if (ch = 'Z' + 1) {ch = 'A' ;}// enter the row for (I = 1; I <n; I ++) {r [x] [y] = ch; y --; ch ++; if (ch = 'Z' + 1) {ch = 'A ';}} // fill in the column for (I = 1; I <m; I ++) {r [x] [y] = ch; x --; ch ++; if (ch = 'Z' + 1) {ch = 'a';} // recursive call, fill in solve (x + 1, y + 1, ch, m-2, N-2);} int main () {int n; while (scanf ("% d", & n )! = EOF) {solve (, 'A', n-1, n); // print for (int I = 0; I <n-1; I ++) {for (int j = 0; j <n; j ++) {printf ("% c", r [I] [j]);} printf ("\ n") ;}} return 0 ;};