C中多个数组的笛卡尔积

概述我能够在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.有人可以解释一下如何“只使用数组”.如果不可能的话有阵列请告诉我其他解决方案.谢谢.这是我的代码,下面是3个阵列的笛卡尔积. #include<stdio.h>int main(void){ int array1[100]; int array2[100]; in 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.有人可以解释一下如何“只使用数组”.如果不可能的话有阵列请告诉我其他解决方案.谢谢.这是我的代码,下面是3个阵列的笛卡尔积.

#include<stdio.h>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<cartesian_arr_col_len;j++)                {                        if(j==0)                        {                                cartesian_array[i][j]=array1[p/x];       p++;                        }                        if(j==1)                        {                                cartesian_array[i][j]=array2[q/y];                                q++;                                if(q>=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;}

解决方法 你能阅读Java代码并自己翻译吗？

import java.util.*;    class CartesianIterator <T> implements Iterator <List <T>> {        private final List <List <T>> lilio;            private int current = 0;        private final long last;        public CartesianIterator (final List <List <T>> llo) {                lilio = llo;                long product = 1L;                for (List <T> lio: lilio)                        product *= lio.size ();                last = product;        }         public boolean hasNext () {                return current != last;        }        public List <T> next () {                ++current;                return get (current - 1,lilio);        }        public voID remove () {                ++current;        }        private List<T> get (final int n,final List <List <T>> lili) {                switch (lili.size ())                {                        case 0: return new ArrayList <T> (); // no break past return;                        default: {                                List <T> inner = lili.get (0);                                List <T> lo = new ArrayList <T> ();                                lo.add (inner.get (n % inner.size ()));                                lo.addAll (get (n / inner.size (),lili.subList (1,lili.size ())));                                return lo;                        }                }        }}class CartesianIterable <T> implements Iterable <List <T>> {        private List <List <T>> lilio;          public CartesianIterable (List <List <T>> llo) {                lilio = llo;        }        public Iterator <List <T>> iterator () {                return new CartesianIterator <T> (lilio);        }}class CartesianIteratorTest {        public static voID main (String[] args) {                List <Character> la = Arrays.asList (new Character [] {'a','b'});                List <Character> lb = Arrays.asList (new Character [] {'b','c'});                      List <Character> lc = Arrays.asList (new Character [] {'c','a'});                List <List <Character>> llc = new ArrayList <List <Character>> ();                llc.add (la);                llc.add (lb);                llc.add (lc);                CartesianIterable <Character> ci = new CartesianIterable <Character> (llc);                for (List<Character> 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 <List <Character>> ();                llc.add (la);                llc.add (lb);                llc.add (lc);                ci = new CartesianIterable <Character> (llc);                for (List<Character> lo: ci)                        show (lo);        }        public static voID show (List <Character> 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.迭代很可能对你没用.

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一个索引,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 => (1,3)[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

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