C中多个数组的笛卡尔积
我能够在C.中实现静态数组数的笛卡尔积。但是我想构建一个动态地获取输入数组的代码。有人可以解释如何“只使用数组”这样做。如果不可能的话有arrays请告诉我其他解决方案。谢谢。这是我的代码，下面是3个arrays的笛卡尔积。
#include int main(void) { int array1[100]; int array2[100]; int array3[100]; int cartesian_array[100][100]; int m,n,o; int i,j; int p=0,q=0,r=0; int x,y,z; int cartesian_arr_row_len; int cartesian_arr_col_len; printf("Enter the size of first array:\n"); scanf("%d",&m); printf("Enter the size of second array:\n"); scanf("%d",&n); printf("Enter the size of third array:\n"); scanf("%d",&o); printf("Enter the first array elements:\n"); for(i=0;i<m;i++) scanf("%d",&array1[i]); printf("Enter the second array elements:"); for(i=0;i<n;i++) scanf("%d",&array2[i]); printf("Enter the third array elements:"); for(i=0;i<o;i++) scanf("%d",&array3[i]); cartesian_arr_row_len=m*n*o; cartesian_arr_col_len=3; x=cartesian_arr_row_len/m; y=cartesian_arr_row_len/(m*n); z=o; for(i=0;i<cartesian_arr_row_len;i++) { for(j=0;j=n*y) q=0; } if(j==2) { cartesian_array[i][j]=array3[r%z]; r++; } } } printf("The Cartesian Product of two arrays is:\n"); for(i=0;i<cartesian_arr_row_len;i++) { for(j=0;j<cartesian_arr_col_len;j++) { printf("%d\t",cartesian_array[i][j]); } printf("\n"); } return 0; } 你能阅读Java代码并自己翻译吗？
import java.util.*; class CartesianIterator implements Iterator > { private final List
> lilio; private int current = 0; private final long last; public CartesianIterator (final List
> llo) { lilio = llo; long product = 1L; for (List lio: lilio) product *= lio.size (); last = product; } public boolean hasNext () { return current != last; } public List next () { ++current; return get (current - 1, lilio); } public void remove () { ++current; } private List get (final int n, final List > lili) { switch (lili.size ()) { case 0: return new ArrayList (); // no break past return; default: { List inner = lili.get (0); List lo = new ArrayList (); lo.add (inner.get (n % inner.size ())); lo.addAll (get (n / inner.size (), lili.subList (1, lili.size ()))); return lo; } } } } class CartesianIterable implements Iterable > { private List
> lilio; public CartesianIterable (List
> llo) { lilio = llo; } public Iterator
> iterator () { return new CartesianIterator (lilio); } } class CartesianIteratorTest { public static void main (String[] args) { List la = Arrays.asList (new Character [] {'a', 'b'}); List lb = Arrays.asList (new Character [] {'b', 'c'}); List lc = Arrays.asList (new Character [] {'c', 'a'}); List > llc = new ArrayList
> (); llc.add (la); llc.add (lb); llc.add (lc); CartesianIterable ci = new CartesianIterable (llc); for (List lo: ci) show (lo); la = Arrays.asList (new Character [] {'x', 'y', 'z'}); lb = Arrays.asList (new Character [] {'b'}); lc = Arrays.asList (new Character [] {'c'}); llc = new ArrayList > (); llc.add (la); llc.add (lb); llc.add (lc); ci = new CartesianIterable (llc); for (List lo: ci) show (lo); } public static void show (List lo) { System.out.print ("("); for (Object o: lo) System.out.print (o); System.out.println (")"); } }
我不知道C中是否存在与迭代器类似的东西。 iterable很可能对你没用。
代码的想法是，有一个计数器，它在结果集的所有元素上创建索引，并计算绑定到该值的元素。
如果我们有3组（1,2,3）（4,5）（6,7,8），我们有18 = 3x2x3的结果。 我们可以，例如迭代器位置7计算结果如下：
7 % 3 = 1 => (1,2,3)[1] = 2 (number modulo 1st group size) 7 / 3 = 2 (int division) (number div 1st group size) 2 % 2 = 0 => (4,5)[0] = 4 (rest modulo 2nd group size) 2 / 2 = 0 0 % 3 = 0 => (7,8,9) => 7 idx g1 g2 g3 0 1 4 6 1 2 4 6 2 3 4 6 3 1 5 6 4 2 5 6 5 3 5 6 6 1 4 7 7 2 4 7 8 3 4 7 9 1 5 7 10 2 5 7 11 3 5 7 12 1 4 8 13 2 4 8 14 3 4 8 15 1 5 8 16 2 5 8 17 3 5 8