编写poweret代码的问题 [英] Problems with writing powerset code

问题描述

我试图生成一个集的powerset，我写了这个代码。问题是，当用户输入两个类似的成员，它不能正常工作。我能做什么？
这是我的代码：

  #include< iostream> 
 #include< string.h> 
 #include< stdio.h> 
 #include< stdlib.h> 
 #include< math.h> 
 using namespace std; 
 
 char获取（char * p，int n）
 {
 int i; 
 for（i = 0; i  {
 cout<<enter member<<（i + 1）< ; 
 cin>> *（p + i）; 
} 
 return * p; 
} 
 
 void set_display（char * p，int n）
 {
 cout< 
 for（int i = 0; 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  {
 cout< 
 for（j = 0; j  {
 if（i&（1≤j））
 cout < j）。 
} 
 cout<<} \\\
; 
} 
} 
  

解决方案

我无法调试您缺少的代码。但是因为你发布了你的问题我写了一个代码在 C 。可能你觉得有帮助。

 ＃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  if（（i& ; j）））{
 printf（％c，y [j]）; 
} 
} 
 printf（\\\
）; 
} 
 printf（\\\
）; 
 return EXIT_SUCCESS; 
} 
  

其工作原理：

 ：〜$ gcc xc -lm -Wall 
：〜$ ./a.out 
 
a 
b 
ab 
c 
ac 
bc 
abc 
  
 
 
 
   
 
 
 
您的错误大小：两个符号相同时。 
  y [0] ='a'; 
 y [1] ='a'; 
 y [2] ='c'; 
 
：〜$ ./a.out 
 
a 
a 
aa 
c 
ac 
ac 
aac 
  
但是根据集合理论，这是错误的。因为在集合中，我们不能有两次相同的元素（超过onec）。但也输入错误：（a，a，c）不是一个集合。所以代码是可用的。 
 
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";
    }
}

解决方案
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. 

                        
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