用C#输出下面图形,高手过来,看看谁的方法最牛! 1 121 12321 1234321123454321 1234321 12321 121 1 解决方案 » 免费领取超大流量手机卡,每月29元包185G流量+100分钟通话, 中国电信官方发货 学Basic时候的题目,楼主还不会。 用两个for循环嵌套,代码最短 凑个热闹. MessageBox.Show(" 1 \n 121 \n 12321 \n 1234321 \n123454321\n 1234321 \n 12321 \n 121 \n 1 \n";);运行效率最高. class T{static void Main(){System.Console.Write(@" 1 121 12321 1234321 123454321 1234321 12321 121 1";}} 参考如下代码:private void Print(string left, string right, int index, int count){ if (index == count) Console.WriteLine("{0}{1}{2}", left, index, right); else { Print(string.Format("{0," + (count - index) + "}{1}", "", left), right, index, index); Print(left + index, index + right, index + 1, count); Print(string.Format("{0," + (count - index) + "}{1}", "", left), right, index, index); }}private void button1_Click(object sender, EventArgs e){ Print("", "", 1, 5); Print("", "", 1, 9);} 低价转让:中#美#爱#梯#科技.Net软件工程师培训课程内容(共35G),声音清晰,课程完整!需要这个培训视频的朋友请联系 QQ 936652114 不能光说不练,我也试试。public void Pagoda(int tier) { //用等差数列法 int b1 = -1*tier+2; //起始值 int e = 1; //等差数列的公差 int m = 1; //项数 int bm = 0; //第n项的值 for (int i = 1; i <= 2 * tier - 1; i++) { int a1; //起始值 int d = 1; //等差数列的公差 int n = 1; //项数 int an = 0; //第n项的值 m = i; //根据位置将数列变为递减 if (i == tier + 1) { b1 = bm; e = -1; } if (m > tier) { m = i - tier + 1; } bm = b1 + (m - 1) * e * (1); a1 = bm; //输出行 for (int j = 1; j <= 2 * tier - 1; j++) { n = j; //根据位置将数列变为递减 if (j == tier + 1) { a1 = an; d = -1; } if (n > tier) { n = j - tier + 1; } an = a1 + (n - 1) * d * (1); if (an <= 0) { System.Console.Write(" "); } else { System.Console.Write(an); } } System.Console.WriteLine();//换行 } }当参数为5的结果如下 1 121 12321 1234321 123454321 1234321 12321 121 1 select语句 Socket通信中数据覆盖解决方法 数据集xsd里的dataset 绑定到textbox Spring.Context.Support.ContextRegistry”的类型初始值设定项引发异常 汇编学的好的兄弟帮偶看道简单的题目 如何在dataset分组循环处理呢? 这个问题有谁懂,进来看一下谢谢 在C#中怎么将字符串日期转换成日期型呀?? ListView控件 我要在class里调用相对路径,如何实现? 关于在数据集中查询并修改数据的一个问题 能不能锁定注册表?
MessageBox.Show(" 1 \n 121 \n 12321 \n 1234321 \n123454321\n 1234321 \n 12321 \n 121 \n 1 \n";
);运行效率最高.
121
12321
1234321
123454321
1234321
12321
121
1";}}
private void Print(string left, string right, int index, int count)
{
if (index == count)
Console.WriteLine("{0}{1}{2}", left, index, right);
else
{
Print(string.Format("{0," + (count - index) + "}{1}", "", left),
right, index, index);
Print(left + index, index + right, index + 1, count);
Print(string.Format("{0," + (count - index) + "}{1}", "", left),
right, index, index);
}
}private void button1_Click(object sender, EventArgs e)
{
Print("", "", 1, 5);
Print("", "", 1, 9);
}
需要这个培训视频的朋友请联系 QQ 936652114
{
//用等差数列法
int b1 = -1*tier+2; //起始值
int e = 1; //等差数列的公差
int m = 1; //项数
int bm = 0; //第n项的值 for (int i = 1; i <= 2 * tier - 1; i++)
{
int a1; //起始值
int d = 1; //等差数列的公差
int n = 1; //项数
int an = 0; //第n项的值 m = i;
//根据位置将数列变为递减
if (i == tier + 1)
{
b1 = bm;
e = -1;
}
if (m > tier)
{
m = i - tier + 1;
} bm = b1 + (m - 1) * e * (1);
a1 = bm;
//输出行
for (int j = 1; j <= 2 * tier - 1; j++)
{
n = j;
//根据位置将数列变为递减
if (j == tier + 1)
{
a1 = an;
d = -1;
}
if (n > tier)
{
n = j - tier + 1;
} an = a1 + (n - 1) * d * (1);
if (an <= 0)
{
System.Console.Write(" ");
}
else
{
System.Console.Write(an);
}
}
System.Console.WriteLine();//换行
}
}当参数为5的结果如下
1
121
12321
1234321
123454321
1234321
12321
121
1