C + + Calculates the number of all combinations of a number

Calculates the combination number of a number and uses recursion to solve it.

If you calculate the number of 3-bit combinations, first select a fixed digit, and then calculate the remaining two-bit combinations, and finally combine them. such as 1 + [23, 32] = 123, 132;

In fixed remaining digits, such as 2 + [13, 31] = 213, 231; 3 + [12, 21] = 312, 321;

The program is divided into two steps, one is to delete an element of any position, one is the number of recursive solution combinations.

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Code:

* * * Combination.cpp * * Created on:2014.6.9 * author:spike//*eclipse CDT, gcc 4.8.1*/  
      
#include <iostream> #include <vector> #include <string> using namespace std;  
    void Deleteonenum (std::string& _num, std::size_t _n) {if (_n >= _num.length ()) {return;  
    A String Temp (_num.substr (_n+1));  
_num = _num.substr (0, _n) + temp;  
} void combination (std::string _num, std::string _buff, std::vector<std::string>& _result)  
    {if (_num.length () <= 0) {_result.push_back (_buff);  
        for (std::size_t i=0; I<_num.length (); ++i) {std::string temp (_num);  
        Deleteonenum (temp, i);  
    Combination (temp, _buff+_num[i], _result);  
    int main (void) {std::string num ("4123");  
    std::vector<std::string> result;  
    Combination (num, "", result); for (Std::size_t i=0; I<result.size ();  
    ++i) {std::cout << result[i] << Std::endl;  
return 0; }

Output:

4123  
4132  
4213  
4231  
4312  
4321  
1423 1432 1243 1234 1342 1324  
2413  
2431  
2143  
2134  
2341 2314 3412 3421 3142  
3124  
3241  
3214

Author: csdn Blog spike_king