回溯法之求n个集合的幂集

void GetPowerSet(int i,int length,int A[],int B[]){
int C
;
memcpy(C, B, 10 * sizeof(int));  //复制数组B,解决c++数组的传址的问题

if (i >= length){
printfArr1(C, length);  //打印出幂集的一个元素
}
else
{
GetPowerSet(i + 1, length, A, C);       //不选择第i个元素，直接处理数组的第i+1个元素
C[i] = A[i];                        //选择第i个元素
GetPowerSet(i+1,length,A,C);
}
}

void printfArr1(int B[],int length){
if (length > 0){

for (int i = 0; i<length; i++){
if (i + 1 == length)
{
if (B[i] == 0){printf("\n"); continue;}
printf("%d \n", B[i]);
}
else {
if (B[i] == 0)continue;
printf("%d ,", B[i]);
}
}
}
}

const int N=10;
int main()
{
int A
= { 2, 1, 3, 4, 5, 6, 7, 8, 9, 10 };
int D[N] = { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 };
GetPowerSet(0, N, A, D);
return 0;
}