/*知道什么是魔幻矩阵吗？它是这样的一个矩阵：首先它是一个奇数阶矩阵；其次无论它的每一行之和还是每一列之和还是左对角线之和还是右对角线之和都等于一个相同的数。神奇吧？那么这样的一个魔幻矩阵是怎么构成的呢？下面这个小程序就帮你解决这个问题。为了便于理解，代码中有些冗余。你可以在理解了源代码之后再对它进行求精。别看程序不起眼，可是很人性化呢！运行一下，看看效果*********************/
#include <iostream.h>
#include <stdio.h>
#define N 10
void main()
{
int i,j;
int n,m;
int left,right;
int a[N][N];
char ch1,ch2;
L:
do{
  cout<<"Please enter a positive number between(1-"<<N-1<<"):";
  cin>>m;
  if(m<1||m>N-1)
       cout<<"Invalid number!Again!"<<endl;
  if(m%2!=1)
   cout<<"Are You Sure it is an odd number?Aain!"<<endl;
}while((m<1||m>N-1)||m%2!=1);
for(i=0;i<N;i++)
  for(j=0;j<N;j++)
   a[i][j]=0;
i=0;
j=m/2;
n=1;
a[i][j]=1;
while(n<m*m)
{
  i--;
  j++;
  n++;
  if((i<0&&j<m)&&a[m-1][j]==0)
  {a[m-1][j]=n;
    i=m-1;
j=j;
  }
  else
     if((i<m-1&&j>m-1)&&a[i][0]==0)
  { a[i][0]=n;
     i=i;
  j=0;
  }
  else
        if(a[i][j]!=0&&a[i+2][j-1]==0)
  {a[i+2][j-1]=n;
    i=i+2;
    j=j-1;
  }
  else
     if(i<0&&j>m-1)
     {a[i+2][j-1]=n;
        i=i+2;
     j=j-1;
     }
     else
      a[i][j]=n;
}
cout<<"Magic Array:"<<endl;
for(i=0;i<m;i++)
  for(j=0;j<m;j++)
   a[i][m]+=a[i][j];
for(j=0;j<m;j++)
  for(i=0;i<m;i++)
   a[m][j]+=a[i][j];
for(i=0,j=0;i<m,j<m;i++,j++)
  a[m][m]+=a[i][j];
left=0;
for(i=0,j=m-1;i<m,j>=0;i++,j--)
  left+=a[i][j];
right=a[m][m];
for(i=0;i<m+1;i++)
{for(j=0;j<m+1;j++)
   printf("%4d",a[i][j]);
  cout<<endl;
}
cout<<"left="<<left<<endl;
cout<<"right="<<right<<endl;
do{
  cout<<"GO ON?(Y/N)";
  cin>>ch1;
  if((ch1!='Y'&&ch1!='y')&&(ch1!='N'&&ch1!='n'))
   cout<<"Invalid char!Again!"<<endl;
}while((ch1!='Y'&&ch1!='y')&&(ch1!='N'&&ch1!='n'));
if(ch1=='Y'||ch1=='y')
  goto L;
else
{
   do{
    cout<<"Really want to EXIT?(Y/N)";
    cin>>ch2;
    if((ch2!='Y'&&ch2!='y')&&(ch2!='N'&&ch2!='n'))
   cout<<"Invalid char!Again!"<<endl;
   }while((ch2!='Y'&&ch2!='y')&&(ch2!='N'&&ch2!='n'));
   if(ch2=='N'||ch2=='n')
    goto L;
   else
    cout<<"Welcome to come back again!"<<endl;
  }
}
/*程序不全由本人书写，转载！VC 6.0调试通过*/