static void Main(string[] args)
{
int i, j, k, n;
n = 4;
for (i = 1; i <=n; i++)
{
for (k = 1; k <=n -1; k++)
{
Console.Write(" ");
}
for (j = 1; j <=2*i-1; j++)
{
Console.Write("*");
}
Console.Write("\n");
}
}
我写了上面的代码,想要输出这样一个三角形:
*
* *
* * *
* * * *
{
int i, j, k, n;
n = 4;
for (i = 1; i <=n; i++)
{
for (k = 1; k <=n -1; k++)
{
Console.Write(" ");
}
for (j = 1; j <=2*i-1; j++)
{
Console.Write("*");
}
Console.Write("\n");
}
}
我写了上面的代码,想要输出这样一个三角形:
*
* *
* * *
* * * *
**
***
**** *
***
*****
*******
就是要你第二个三角形
{
int i, j, k, n;
n = 4;
for (i = 1; i <= n; i++)
{
for (k = 1; k <= n - i; k++)
{
Console.Write(" ");
}
for (j = 1; j <=2*i-1; j++)
{
Console.Write("*");
}
Console.Write("\n");
}
}
{
int i, j, k, n;
n = 4;
for (i = 1; i <=n; i++)
{
for (k = 1; k <=n -1; k++)
{
Console.Write("A");
}
for (j = 1; j <=2*i-1; j++)
{
Console.Write("*");
}
Console.Write("\n");
}
for(i=0;i<=m-1;i++)
{for(j=0;j<=(m-1)-i;j++)
Response.Write(" ");
for(k=0;k<=i;k++)
Response.Write("* ");
Response.Write("<br/>");
}
***
*****
*******用 [code=C#]...[/code] 就可以,把其中全角的[]改成半角的[]。
print( 10 );
Console.ReadLine();
} static void print( int i ) {
if( i >= 0 ) {
int j = 0;
while( j < i ) {
Console.Write( " *" );
j++;
}
Console.WriteLine();
print( --i );
}
}
不过星星不居中...
{
static void Main(string[] args)
{
int i, j, k, n;
n = 4;
for (i = 1; i <= n; i++)
{
for (k = 1; k <= n - i; k++)
{
Console.Write(" ");
}
for (j = 1; j <= 2*i-1; j++)
{
Console.Write(j % 2 == 0 ? " " : "*");
}
Console.Write("\n");
}
} }
/* 输出: *
* *
* * *
* * * **/
{
static void Main()
{
int n = 4;
for (int i = 1; i <= n; i++)
{
Console.Write(new string(' ', n - i));
for (int j = 1; j <= 2 * i - 1; j++)
Console.Write(" *"[j % 2]);
Console.WriteLine();
}
}
}/* 输出: *
* *
* * *
* * * **/
对于n层的三角形.在第x层星号数S = f(x)= 2*x-1 (0<x<+∞|x∈N)
第x层输出的空格数 W = g(x) = (int(n/2) + 1) - x (0<x<+∞|x∈N)