使用递归。
设数组a1 a2 ... an,
定义两个数组联合运算为"+"
则
a1+a2+a3+...+an=a1+(a2+a3+...+an)=a1+(a2+(a3+(...+(an-1+an)))))
设数组a1 a2 ... an,
定义两个数组联合运算为"+"
则
a1+a2+a3+...+an=a1+(a2+a3+...+an)=a1+(a2+(a3+(...+(an-1+an)))))
调试欢乐多
using System.Collections;
class arr
{
static string[] a=new string[] {"1","2","3"};
static string[] b=new string[] {"a","b","c"};
static string[] c=new string[] {"A","B","C"};
static string[] d=new string[a.Length*b.Length*c.Length];
static int m=0;
static void Main()
{
getnewarr(a,b,c);
print(d);
} public static string[] getnewarr(string[] a,string[] b,string[] c)
{
for(int i=0;i<a.Length;i++)
{
for(int j=0;j<b.Length;j++)
{
for(int k=0;k<c.Length;k++)
{
d[m]=a[i]+b[j]+c[k];
m++;
}
}
}
return d;
}
static void print(string[] n)
{
for(int f=0;f<n.Length;f++)
{
Console.WriteLine(n[f]);
}
}
}