#include <iostream>
using namespace std;
void Output(unsigned a[],unsigned k,unsigned m,unsigned n) //k表示此次输出的个数,m表示从第几个开始输出
{
for(unsigned j=k+m;j<n;j++)
{ cout<<"<";
unsigned t=a[j];a[j]=a[k+m];a[k+m]=t;
for(unsigned i=m;i<=k+m;i++)
cout<<a[i];
t=a[j];a[j]=a[k+m];a[k+m]=t;
cout<<">";
}
}
void Powerset(unsigned a[],unsigned n)
{
cout<<"<>";
for(unsigned i=0;i<n;i++)
cout<<"<"<<a[i]<<">";
for(unsigned i=1;i<n;i++)
{
for(unsigned j=0;j<=n;j++)
{
Output(a,i,j,n);
}
}
}
int main()
{
unsigned a[]={1,2,3,4,5,6};
Powerset(a,5);
}
非递归求集合的幂集
最新推荐文章于 2024-01-27 18:36:19 发布
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