操作数的结合率

[shancao]

乘法具有结合率
a*b*c有2种结合方法。a*b*c=(a*b)*c=a*(b*c)
a*b*c*d=(a*b)*(c*d)=((a*b)*c)*d=(a*(b*c))*d=a*(b*(c*d))=a*((b*c)*d)
编程实现n个操作数有多少种结合方法？
谢谢啦

[wargrey]

思路，递归实现


a*b*c*d...*z=a*B(B=b*c*d...*z),出口条件是只到B为一个成数的情况

[wjb_yd]

动态规划

[K行天下]

做了一个：

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

set<string> multiNum(int n)
{
    if ( n < 2)
    {
        cerr<<"乘法个数小于2"<<endl;
    }
    else if ( n == 2 )
    {
        set<string> s;
        s.insert("X*X");
        return s;
    }
    else if( n > 2)
    {
        set<string> sn1 = multiNum(n-1);
        set<string> s;
        set<string>::iterator sitern1 = sn1.begin();

        while(sitern1!=sn1.end())
        {
            string str = *sitern1;
            cout<<"string str = *sitern1  "<<str<<endl;
            cout<<str.size();
            for(int i=0;i<str.size();i++)
            {
                if(str[i] == 'X')
                {
                    string temp;
                    cout<<endl<<"遇见X"<<temp<<endl<<endl;
                    temp = string(str,i-1)+string("(X*X)")+string(str,i+1,str.size()-i-1);
                    s.insert(temp);

                }
            }
        }
         return s;
    }
}

int main()
{
    set<string> s = multiNum(2);
    set<string>::iterator siter = s.begin();
    while(siter != s.end())
    {
        cout<<*siter<<endl;//最后转换输出
        siter++;
    }
    
    multiNum(3);
    cin.get();
    return 0;
}

[K行天下]

前面那个全部用 X 表示的，改为以下形式：

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

set<string> multiNum(int n)
{
    if ( n < 2)
    {
        cerr<<"乘法个数小于2"<<endl;
    }
    else if ( n == 2 )
    {
        set<string> s;
        s.insert("X*X");
        return s;
    }
    else if( n > 2)
    {
        set<string> sn1 = multiNum(n-1);
        set<string> s;
        set<string>::iterator sitern1 = sn1.begin();

        while(sitern1 != sn1.end())
        {
            string str = *sitern1;
            for(int i=0;i<str.size();i++)
            {
                if(str[i] == 'X')
                {
                    string temp;
                    if( i == 0)
                        temp = string("(X*X)")+string(str,i+1,str.size()-i-1);
                    else if( i == str.size()-1)
                        temp = string(str,0,i)+string("(X*X)");
                    else
                        temp = string(str,0,i)+string("(X*X)")+string(str,i+1,str.size()-i-1);
                    s.insert(temp);
                }
            }
            sitern1++;
        }
         return s;
    }
}


void turn2alpha(set<string> s)
{
    set<string>::iterator siter = s.begin();
    while(siter!=s.end())
    {
        string s = *siter;
        char* myAlpha = new char[2];
        myAlpha[0] = 'a';
        myAlpha[1] = '\0';
        for(int i=0; i<s.size(); i++)
        {
            if(s[i] == 'X')
            {
                s.replace(i,1,myAlpha);
                ++*myAlpha;
            }
        }
        delete[] myAlpha;
        cout<<s<<endl;
        siter++;
    }

}
int main()
{
    for(int i =2; i<6; i++)
    {
        /*
        set<string> s = multiNum(i);
        set<string>::iterator siter = s.begin();

        while(siter != s.end())
        {
            cout<<*siter<<endl;//最后转换输出
            siter++;
        }*/
        cout<<i<<"个乘数的所有结合律如下："<<endl;
        turn2alpha(multiNum(i));
    }

    cin.get();

    cout<<"10个乘数的所有结合律如下："<<endl;
        turn2alpha(multiNum(10));

    cin.get();
    return 0;
}