Listen Sie bei gegebener Zeichenfolge alle möglichen Längenzeichenfolgen auf, aus denen sie besteht

Begin
   n := length of the string str
   count is 2^n – 1
   for each number 0 to count, do
      sub_str := empty string
      for j in range 0 to n, do
         if jth bit of the counter is set, then
            concatenate jth character of str with sub_str
         end if
      done
      repeat:
         print sub_string
      until next permutation of sub_string is not completed
   done
End

#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
void printAllString(string str) {
   int n = str.size();
   unsigned int count = pow(2, n);
   for (int counter = 1; counter <count; counter++) { //generate 2^n - 1 strings
      string subs = "";
      for (int j = 0; j < n; j++) {
         if (counter & (1<<j)) //when the jth bit is set, then add jth character
            subs.push_back(str[j]);
      }
      do{
         cout << subs << endl;
      }
      while (next_permutation(subs.begin(), subs.end()));
   }
}

A
B
AB
BA
C
AC
CA
BC
CB
ABC
ACB
BAC
BCA
CAB
CBA
D
AD
DA
BD
DB
ABD
ADB
BAD
BDA
DAB
DBA
CD
DC
ACD
ADC
CAD
CDA
DAC
DCA
BCD
BDC
CBD
CDB
DBC
DCB
ABCD
ABDC
ACBD
ACDB
ADBC
ADCB
BACD
BADC
BCAD
BCDA
BDAC
BDCA
CABD
CADB
CBAD
CBDA
CDAB
CDBA
DABC
DACB
DBAC
DBCA
DCAB
DCBA