素數幻方：4*4矩陣中，每格填入一個數字，使每一行、每一列和兩條對角線上的4個數字所組成的四位數，均爲可逆素數。

/*************************************************************************
    > File Name: 35.c
    > Author: ma6174
    > Mail: [email protected] 
    > Created Time: 2018年07月29日 星期日 19時32分03秒
 ************************************************************************/

#include <stdio.h>
#include <math.h>

int number[210][5];
int select[110];
int array[4][5];
int count;
int selecount;
int larray[2][200];
int lcount[2];
	
int main()
{
	int i, k, flag, cc = 0, i1, i4;
	printf("there are magic squares with invertable primes as follow:\n");
	for ( i = 1001; i < 9999; i += 2 )
	{
		k = i/1000;
		if( k % 2 != 0 && k != 5 && num(i))
		{
			number[count][0] = i;
			process(count ++);
			if(number[count -1][2]%2 != 0 &&
			   number[count -1][3]%2 != 0 &&
			   number[count -1][2] != 5 &&
			   number[count -1][3] != 5)
			   {
				   select[selecount++] = count - 1;
			   }
		}
	}
	larray[0][lcount[0]++] = number[0][0] / 100;
	larray[1][lcount[1]++] = number[0][0] / 10;
	
	for ( i = 1; i < count; i++)
	{
		if(larray[0][lcount[0]-1] != number[i][0] / 100)
		{
			larray[0][lcount[0]++] = number[i][0] / 100;
		}
		if(larray[1][lcount[1]-1] != number[i][0] / 10)
		{
			larray[1][lcount[1]++] = number[i][0] / 10;
		}
	}
	for (i1 = 0; i1 < selecount; i1++)
	{
		array[0][0] = select[i1];
		copy_num(0);
		
		for (array[1][0] = 0; array[1][0] < count; array[1][0]++)
		{
			copy_num(1);
			if(!comp_num(2))
			{
				continue;
			}
				for (array[2][0] = 0; array[2][0] < count; array[2][0]++)
				{
					copy_num(2);
			
					if( !comp_num(3) )
					{
						continue;
					}
						for(i4 = 0; i4 < selecount; i4++)
						{
							array[3][0] = select[i4];
								copy_num(3);
								for(flag = 1, i = 1; flag && i <= 4; i++)
								{
									if( !find1(i))
									{
										flag = 0;
									}
									if( flag && find2())
									{
										printf("No.%d\n", ++cc);
										p_array();
									}
								}
			            }
		        }			
	    }   
    }
return 0;
}

num (int number)
{
	int j;
	if( !ok(number))
	{
		return(0);
	}
	for(j = 0;number>0;number/=10)
	{
		j = j*10 + number%10;
	}
	if( !ok(j))
	{
		return(0);
	}
	return 1;
}
	
ok(int number)
{
	int i,j;
	if( number%2 == 0)
	{
		return(0);
	}
	j = sqrt((double)number) +1;
	for( i = 3;i <= j; i += 2)
	{
		if(number % i == 0)
		{
			return(0);
		}
	}
	return(1);
}
process(int i)
{
	int j,num;
	num = number[i][0];
	for( j =4; j>=1;j--,num /= 10)
	{
		number[i][j] = num %10;
	}
}

copy_num(int i)
{
	int j;
	for( j = 1;j <= 4;j++)
	{
		array[i][j] = number[array[i][0]][j];
	}
}

comp_num(int n)
{
	static int ii;
	static int jj;
	int i,num,k,*p;
	int *pcount;
	switch(n)
	{
		case 2:
				pcount = & lcount[0]; p = &ii;
				break;
		case 3:
				pcount = & lcount[1]; p = &jj;
				break;
		default: 
				return 0;
	}
	for (i = 1; i <= 4; i++)
	{
		for(num = 0,k= 0; k<n;k++)
		{
			num = num * 10 + array[k][i];
		}
		if (num <= larray[n-2][*p])
		{
			for( ; *p>=0 && num< larray[n-2][*p];(*p)--)
			{
				;
			}
		}
		else
		{
			for( ; *p< *pcount && num >larray [n-2][*p]; (*p)++)
			{
				;
			}
		}
		if( *p<0 || *p >= *pcount)
		{
			*p = 0;
			return 0;
		}
		if(num != larray[n-2][*p])
		{
			return 0;
		}
	}
	return 1;
}
	
find1(int i)
{
	int num,j;
	for(num = 0, j = 0; j < 4; j++)
	{
		num = num * 10+array[j][i];
	}
	return(find0(num));
}
	
find2(void)
{
	int num1, num2, j, i;
	for ( num1 = 0, j = 0; j < 4; j++)
	{
		num1 = num1 * 10 + array[j][j+1];

		for ( num2 = 0, j = 0, i = 4; j < 4; j++, i--)
		
			num2 = num2 * 10 + array[j][i];
		
		if( find0(num1)) 
		
			return (find0(num2));
		
		else 
			return 0;
	}
	
find0(int num)
{
	static int j;
	if(num <= number[j][0])
	{
		for(; j>=0 && num< number[j][0]; j--)
		{
			;
		}
	}
	else 
		for( ; j<count && num > number[j][0]; j++)
		{
			;
		}
	if(j<0 || j >= count)
		{
			j=0;
			return 0;
		}
		if(num == number[j][0])
		{
			return 1;
		}
		else 
			return 0;
	}
	
	p_array(void)
	{
		int i,j;
		for (i = 0; i <= 4; i++)
		{
			for (j = 1; j <= 4; j++)
			{
				printf("%6d", array[i][j]);
			}
			printf("\n");
		}
	}