面向对象程序设计（基于C++）0301字符组合

题目

原题请参见1039. 字符组合
但是这次要求用set容器装好容器后再依次输出

代码模板

/***************************************************************/
/*                                                                                     */
/*  DON'T MODIFY main function ANYWAY!                     */
/*                                                                                      */
/***************************************************************/

#include <iostream>
#include <string>
#include <set>
using namespace std;

//********** Specification of Combination **********
set<string> Combination(string s);
/* PreCondition:
     s is a string with distinct characters in ascending order
   PostCondition:
    return  combination of string s
*/

/***************************************************************/



int main(int argc, char** argv)
{
    string s;
    cout << "Enter a string: ";
    cin >> s;

    cout << endl << "Result: "<< endl;
    bool first{true};

//********** Combination is called here **************
    for (const auto& x: Combination(s))
//**************************************************
    {
        if (first) first=false;
        else cout << " ";
        cout << x;
    }
    cout << endl;
    return 0;
}

思路

dfs在1039中有提及，这里由于需要用到set容器

题目提示用递归来写。。。学术不精。没想出来一个参数怎么弄递归

所以用迭代，直接二进制枚举符合的所有情况，然后一股脑塞进set让他自己排序

代码

/***************************************************************/
/*                                                                                     */
/*  DON'T MODIFY main function ANYWAY!                     */
/*                                                                                      */
/***************************************************************/

#include <iostream>
#include <string>
#include <set>
using namespace std;

set<string> SET;
long long power(long long base, long long n)//处理二进制
{
    if(n == 0) return 1;
    long long temp = base;
    for(int i = 1; i < n; i++)
    {
        base *= temp;
    }
    return base;
}
//********** Specification of Combination **********
set<string> Combination(string s)
{
   int size = power(2,s.length());
   while(size--)
   {
       int bin = size;
       string a = "";
       int loc = s.length() - 1;
       while(bin)
       {
           if(bin & 1)
           {
               a.insert(0,1,s[loc]);
           }
           loc--;
           bin /= 2;
       }
        SET.insert(a);
   }
   return SET;
};
/* PreCondition:
     s is a string with distinct characters in ascending order
   PostCondition:
    return  combination of string s
*/

/***************************************************************/



int main(int argc, char** argv)
{
    string s;
    cout << "Enter a string: ";
    cin >> s;

    cout << endl << "Result: "<< endl;
    bool first{true};

//********** Combination is called here **************
    for (const auto& x: Combination(s))
//**************************************************
    {
        if (first) first=false;
        else cout << " ";
        cout << x;
    }
    cout << endl;
    return 0;
}