找出帕斯卡钻石项目中空间的数学公式
伙计们,进展如何?
我编写了一个在命令行中“绘制”钻石的程序(这是我作业的一部分)。对于菱形内部的空间,我得到了一个公式“1 + 2(k-2) 或 2k -3 ,其中 k 是行号”,但我不明白这个公式是如何创建的。有人能解释一下吗?
program diamond;
var
n, k, h, i: integer;
begin
repeat
write('Enter the diamond''s height (positive odd): ');
readln(h);
until (h > 0) and (h mod 2 = 1);
n := h div 2;
for k := 1 to n + 1 do
begin
for i := 1 to n + 1 - k do
write(' ');
write('*');
if k > 1 then
begin
for i := 1 to 2*k - 3 do
write(' ');
write('*')
end;
writeln
end;
for k := n downto 1 do
begin
for i := 1 to n + 1 - k do
write(' ');
write('*');
if k > 1 then
begin
for i := 1 to 2*k - 3 do
write(' ');
write('*')
end;
writeln
end
end.
How is it going guys ?
I've written a program that "draws" diamond in the command line (it's a part of my homework). For spaces inside the diamond I was given a formula "1 + 2(k-2) or 2k -3 , where k is line number", but I don't understand how this formula was created. Could anyone explain it ?
program diamond;
var
n, k, h, i: integer;
begin
repeat
write('Enter the diamond''s height (positive odd): ');
readln(h);
until (h > 0) and (h mod 2 = 1);
n := h div 2;
for k := 1 to n + 1 do
begin
for i := 1 to n + 1 - k do
write(' ');
write('*');
if k > 1 then
begin
for i := 1 to 2*k - 3 do
write(' ');
write('*')
end;
writeln
end;
for k := n downto 1 do
begin
for i := 1 to n + 1 - k do
write(' ');
write('*');
if k > 1 then
begin
for i := 1 to 2*k - 3 do
write(' ');
write('*')
end;
writeln
end
end.
如果你对这篇内容有疑问,欢迎到本站社区发帖提问 参与讨论,获取更多帮助,或者扫码二维码加入 Web 技术交流群。
绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱,如果其他用户或者作者回复了您的评论,将不能在第一时间通知您!
发布评论
评论(1)
我已经想通了。这是一个简单但经过修改的算术级数 An=A1-d(n-2)。通常我们会使用(n-1),但因为我们需要从每行减去2颗星(从第二行开始,因为这个公式适用于k>1),所以我们使用(n-2)
I've already figured it out. It's a simple, but modified arithmetic progression An=A1-d(n-2). Usually we would use (n-1), but because we need to substract 2 stars from each line (starting from the second one, as this formula works for k>1), we use (n-2)