打印数组对角线路径数组 [英] print array diagonal routes array
本文介绍了打印数组对角线路径数组的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
如果数组的数量为
{ 1 ,我会将此代码设为打印数组, 2 , 3 , 4 ,< span class =code-digit> 5 ,
6 , 7 , 8 , 9 , 10
, 11 , 12 , 13 , 14 , 15
, 16 , 17 , 18 , 19 , 20
, 21 , 22 , 23 , 24 , 25 }
此代码的输出为1,2,6,3,7,11,4,6,12,16,5,9,13,17,21,10,14,18,21,15 ,19,23,20,24,25
我想改变我的代码是从25 20 24 15 19 23 10 14 18 22 5 9 13 17 21 4 8 12 16 3 7 11 2 6 1
int [,] p = new int [ 5 , 5 跨度>];
int sum = 1 ;
for ( int i = 1 ; i < = 5 ; i ++)
{
for ( int j = 1 ; j < = 5 ; j ++)
{
p [i,j ] =总和;
richTextBox1.Text + = p [i,j] .ToString();
总和++;
}
}
int C;
int R = 1 ;
for ( int i = 1 ; i < = 5 ; i ++)
{
C = i;
for ( int r = 1 ; r < = i; r ++)
{
输出=输出+ p [r, C];
C--;
}
}
richTextBox2.Text + = Output.ToString();
for ( int i = 2 ; i > = 5 ; i ++)
{
R = i;
for (C = 5 ; C > ; = i; C--)
{
输出=输出+ p [R,C];
R ++;
}
}
richTextBox2.Text + = Output.ToString();
解决方案
string 输出= ;
int [,] p = new int [ 5 , 5 ];
int sum = 1 ;
for ( int i = 0 ; i < 5 ; i ++)
{
for ( int j = 0 ; j < 5 ; j ++)
{
p [i,j] = sum;
richTextBox1.Text + = p [i,j] .ToString();
总和++;
}
}
for ( int i = 4 ; i > = 0 ; i--)
{
for ( int C = 0 ; C + i< 5& C< 5; C ++)
{
输出=输出+ + p [i + C,4-C];
}
}
for ( int i = 3 ; i > = 0 ; i--)
{
for ( int j = 0 ; j < 4 && i-j> = 0 ; j ++)
{
输出=输出+ + p [j,i -j];
}
}
richTextBox2.Text + = Output.ToString();
// 如果您对解决方案有任何疑问,请询问
Hi,I make this code to print array if the number of array is
{ 1,2,3,4,5,
6,7,8,9, 10
, 11,12,13,14,15
,16,17,18,19,20
,21,22,23,24,25}
the out put of this code is 1,2,6,3,7,11,4,6,12,16,5,9,13,17,21,10,14,18,21,15,19,23,20,24,25
Iwant to change in this my code to be the out put begging from 25 20 24 15 19 23 10 14 18 22 5 9 13 17 21 4 8 12 16 3 7 11 2 6 1
int [,] p = new int [5,5];
int sum = 1;
for (int i = 1; i <= 5; i++)
{
for (int j = 1; j <= 5; j++)
{
p[i, j] = sum;
richTextBox1.Text += p[i, j].ToString();
sum++;
}
}
int C;
int R=1;
for (int i = 1; i <= 5; i++)
{
C = i;
for (int r = 1; r <= i; r++)
{
Output = Output + p[r, C];
C--;
}
}
richTextBox2.Text += Output.ToString();
for (int i = 2; i >= 5; i++)
{
R = i;
for (C = 5; C >= i; C--)
{
Output = Output + p[R, C];
R++;
}
}
richTextBox2.Text += Output.ToString();
解决方案
string Output = ""; int[,] p = new int[5, 5]; int sum = 1; for (int i = 0; i < 5; i++) { for (int j = 0; j < 5; j++) { p[i, j] = sum; richTextBox1.Text += p[i, j].ToString(); sum++; } } for (int i = 4; i >=0; i--) { for (int C = 0; C+i<5&& C<5; C++) { Output = Output +" "+ p[i+C, 4-C]; } } for (int i = 3; i >= 0; i--) { for (int j = 0; j < 4 &&i-j>=0; j++) { Output = Output + " " + p[j,i -j]; } } richTextBox2.Text += Output.ToString(); //If you have any qustion on the Solution please ask
这篇关于打印数组对角线路径数组的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!
查看全文