问题
I'm trying to generate the powerset of a set and I wrote this code. The problem is, when user enter two similar member of the set it dosen't work properly. What can I do? Here is my code:
#include <iostream>
#include <string.h>
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
using namespace std;
char obtain(char *p,int n)
{
int i;
for(i=0;i<n;i++)
{
cout<<"enter member"<<(i+1)<<"\n";
cin>>*(p+i);
}
return *p;
}
void set_display(char *p,int n)
{
cout<<"{";
for(int i=0;i<n;i++)
{
cout<<*(p+i)<<",";
}
cout<<"}";
}
void powset(char *p,int n)
{
unsigned int m = (double)pow((double)2, n);
int i, j;
for(i = 0; i < m; i++)
{
cout<<"{";
for(j = 0; j < n; j++)
{
if(i& (1<<j))
cout<<*(p+j);
}
cout<<"}\n";
}
}
回答1:
Ok I could not debug your code where you are missing. but since you posted your question I written an code in pure C
. May be you find it helpful.
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
int main(int argc, char* argv[]){
int N;
char y[20]={0};
int i,j;
y[0] = 'a';
y[1] = 'b';
y[2] = 'c';
N = 3;
int powerSet;
powerSet=pow(2,N);
for(i = 0; i<(powerSet);i++){
for(j=0;j<N;j++){
if((i & (1<<j))){
printf("%c ",y[j]);
}
}
printf("\n");
}
printf("\n");
return EXIT_SUCCESS;
}
And its working:
:~$ gcc x.c -lm -Wall
:~$ ./a.out
a
b
a b
c
a c
b c
a b c
[ANSWER]
You error case: When two symbols are same.
y[0] = 'a';
y[1] = 'a';
y[2] = 'c';
:~$ ./a.out
a
a
a a
c
a c
a c
a a c
But this is wrong according to set theory. Because in Set we can't have same element twice (more then onec). BUT ALSO INPUT IS WRONG: (a, a, c)
is not a set. So code is usable.
来源:https://stackoverflow.com/questions/15028670/problems-with-writing-powerset-code