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Getting all number combinations in C ++

I need to get all the possible combinations of this set. For example, given: [1, 4, 7], the resulting combinations should be:

  • 111, 114, 117, 141, 144, 147, 171, 174, 177, 411, 414, 417, 441, 444, 447, 471, 474, 477, 711, 714, 717, 741, 744, 747, 771, 774, 777.

I tried using the method next_permutation, but that is not what I want (it does not return values ​​like 111, 144, 717, etc.).

So what can I do this in C ++? Please note that I am a complete newbie.

Thanks in advance.

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7 answers

: {11,14,17,41,44,47,71,74,77} 1, 4 7. :

3 {1,4,7} 2 .

3 2 , , .

++, , , , . base-10. .

+6

, . {1, 4, 7}.

() , ().

100 * array [i] + 10 * array [j] + array [k]


, , , . :

void Solve(int[] array, int length, int position, int sum)
{
    position++;
    sum *= 10;

    for (int cnt = 0; cnt < length; cnt++)
    {
        int tempsum = sum + array[cnt];

        if (position == length)
            output(tempsum);
        else
            Solve(array, length, position, tempsum);
    }
}
+1

( vector), . 0. :

1) vector, . (, 0,1,1, - [9,8,7], 988 - , - .)

2) element 0.

3) vector[element]. , vector[element] . 1.

4) vector[element] . element. element , 3.

5) . .

+1

: PHP

: ++-, . ( )

0

, . . :

#include <iostream>
#include <vector>
#include <algorithm>
#include <set>
using namespace std;

vector<int> allowedNumbers;

bool my_next_perm(vector<int>::iterator& begin, vector<int>::iterator& end) {
    for (vector<int>::iterator itr = end - 1; ; --itr)
    {
        if (*itr != allowedNumbers.back())
        {
            *itr = *upper_bound(allowedNumbers.begin(), allowedNumbers.end(), *itr);
            while ((++itr) != end) {
                *itr = allowedNumbers[0];
            }
            return true;
        }
        if (itr == begin)
        {
            return false;
        }
    }
}
void printAllPermutationsWithRepetitions(vector<int>& v)
{
    int n = v.size();
    set<int> allNumbers;
    for (int i = 0; i < n; i++)
    {
        allNumbers.insert(v[i]);
    }
    for (set<int>::iterator itr = allNumbers.begin(); itr != allNumbers.end(); ++itr) {
        allowedNumbers.push_back(*itr);
    }
    for (int i = 0; i < n; i++) {
        v[i] = allowedNumbers[0];
    }

    do {
        for (int i = 0; i < n; i++) {
            if (i != 0) cout << " ";
            cout << v[i];
        }
        cout << endl;
    } while(my_next_perm(v.begin(), v.end()));
}

int main (int argc, char *argv[]) {
    vector<int> v;
    v.push_back(7);
    v.push_back(1);
    v.push_back(3);
    printAllPermutationsWithRepetitions(v);
    return 0;
}

( upper_bound).

0

You effectively calculate from base 3 from 000 to 222 by looking at your array [1, 4, 7] with these numbers as indices. Generalizing this to work with an arbitrary size of the input array:

int n = digits.size();
int n_n = std::pow(n, n);
for (int i = 0; i < n_n; ++i)
{
     int x = i;
     for (int j = 0; j < n; ++j)
     {
          cout << digits[x % n];
          x /= n;
     }
     cout << ' ';
}
0
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This is the easiest way to do this. You can take any number as input from a user of type 147 and print all permutations as: 111, 114, 117, 141, 144, 147, 171, 174, 177, 411, 414, 417, 441, 444, 447, 471, 474, 477, 711, 714, 717, 741, 744, 747, 771, 774, 777.

#include <iostream>
#include <cstdlib>
#include <cmath>
using namespace std;

 bool Search(int A[],int num,int length)
    {
        for(int i = 0;i<length;i++)
            if(A[i] == num)
                return 1;
        return 0;
    }


    int main()
    {
        int n;
        cout << "Please Enter a Number\n";
        cin>>n;
        int num = n, k = 1;
        int count = 0;
        while(num>0)
        {
            num/=10;
            count++;
        }
        cout<<"The All Permutations of " <<n<<" are: "<<endl;
        num = n;
        int *A = new int[count];

        for(int i = 0;i<count;i++)
        {
            A[i] =  num%10;
            num/=10;
        }


        int fact = pow(count,count);
        int *Ar = new int[fact];
        int *B = new int[count];
        int value,number = 0;

        for(int i = 0;i<fact;)
        {

            for(int j = 0;j<count;++j)
            {
                value = (rand()%count);

                B[j] = value;
            }

            for(int k= 0;k<count;k++)
                number = number + A[B[k]]*pow(10,k);

            if(Search(Ar,number,fact) == 0)
            {
                cout<<k++<<". "<<number<<endl;
                Ar[i] = number;
                ++i;
            }
            number = 0;

        }

        return 0;
    }
0
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