N个数组里选M个元素

#include <stdio.h>
char* str[] =
{
"abc", "de", "f", "gh"
};

void show(char* result, int curr)
{
    if (curr == 4) puts(result);
    else
    {
        int i;
        for (i = 0; str[curr][i]; ++i)
        {
            result[curr] = str[curr][i];
            show(result, curr+1);
        }
    }
}
int main()
{
    char result[16] = {0};
    show(result, 0);
    return 0;
}

 

 

#include <stdio.h>

char *array[] = {"ab","1","24","3",NULL};
char stack[1000];
char top = -1;

void print_array(char **array){
    char *curr = *array;
    if(curr == NULL){
       printf("%s/n",stack);
       return;
    }
    top++;
    while(*curr){
        /*printf("%c",*curr);*/
        stack[top] = *curr;
        print_array(array+1);
        stack[top] = '/0';
        curr++;

    }
    top--;

}


int main(){
     print_array(array);
     getch();
     return 0;
}

#include <stdio.h>

void p(char **s, int n)
{
        int sk[100], t=0, i;
        while (t >= 0) {
                if (t == n) {
                        for (i=0; i<n; i++)
                                putchar(s[i][sk[i]]);
                        putchar('/n');
                        t--;
                        while (t>=0 && !s[t][sk[t]+1])
                                t--;
                        if (t>=0)
                                sk[t++]++;
                } else
                        sk[t++]=0;
        }
}

int main()
{
        char *s[]={"ab", "cd", "gh"};
        p(s, 3);
}

寻找从N个数组中各取1个元素的全部组合

#include <stdio.h>
int r[16] = {0};
void sumit(int n[][5], int sum,int cur,int col,int row)//sum是和的值，curr是当前数组，col和row是二维数组的行和列
{
    if (cur == row)
    {
            if(r[cur]==sum)
            {
             for(int j=0;j<row;j++)
             printf("%-4d",r[j]);
             printf("/n");
            }
    }
    else
    {
        int i;
        for (i = 0; i<col; ++i)
        {
            r[cur] = n[cur][i]; sumit(n, sum-r[cur],cur+1,col,row);
        }
    }
}
int main()
{
    int n[3][5]={{1,2,3,4,5},{6,7,8,9,10},{11,12,13,14,15}};
    sumit(n,23, 0,5,3);
    getchar();
    return 0;
}