Question

假设我们有以下用于素数分解的脚本

 z=input('enter your number : ');
    for ii=2:z
       s=0;
       while z/ii==floor(z/ii) % check if  z is divisible by ii
           z=z/ii;
           s=s+1;
       end
       if s>0
                str = [num2str(ii) '^' num2str(s) ];
            disp(str) ;      
            % If z = 1, no more divisions are necessary, 
            % thus breaks the loop and quits
            if z == 1
                break
            end
        end

    end

但此代码的输出格式不正确，例如

 >> integer_factorization
    enter your number : 30
    2^1
    3^1
    5^1

我怎么能这样做才能得到

30=2^1*3^1*5^1?

提前致谢

Answer 1

这非常简单：使用fprintf(''); instead打印/显示因子。

z=input('enter your number : ');

ans='' %the final answer string

for ii=2:z
   s=0;
   while z/ii==floor(z/ii) % check if  z is divisible by ii
       z=z/ii;
       s=s+1;
   end
   if s>0
            str = [num2str(ii) '^' num2str(s) ];
        %disp(str) ;  

        strcat(ans,'*',str); %concats the * str as per requirement.

        % If z = 1, no more divisions are necessary, 
        % thus breaks the loop and quits
        if z == 1
            break
        end
    end
end

ans=ans(2:end); % to remove the first * 
fprintf(ans); % can even use the disp() function.

所以，基本上，添加了一个字符串来将因子附加到其中并显示在循环之外的末尾。

Answer 2

您可以简单地创建一个字符串并将您的数字添加到字符串中，最后打印字符串。如下：

 z=input('enter your number : ');
for ii=2:z
   s=0;
   while z/ii==floor(z/ii) % check if  z is divisible by ii
       z=z/ii;
       s=s+1;
   end
   if s>0
        str = str + [num2str(ii) '^' num2str(s) '*' ];%only update
        % If z = 1, no more divisions are necessary, 
        % thus breaks the loop and quits
        if z == 1
            break
        end
    end
    str = str(0,str.size -1) %use proper command to remove the last * from your string result
    disp(str) ; %display the str at the end in one line
end

Answer 3

Litle修改你的代码。

z=input('enter your number : ');
for ii=2:z
   s=0;
   while z/ii==floor(z/ii) % check if  z is divisible by ii
       z=z/ii;
       s=s+1;
   end
   if s>0
            str += [num2str(ii) '^' num2str(s) ];

        % If z = 1, no more divisions are necessary, 
        % thus breaks the loop and quits
        if z == 1
            break
        else 
           str+='*';
        end
    end

end
disp(str);

Answer 4

这可行！只需对代码进行一些更改

wheel

Answer 5

首先感谢大家的支持，这是我的最终解决方案

 z=input('enter your number : ');
  string='';
    for ii=2:z
       s=0;
       while z/ii==floor(z/ii) % check if  z is divisible by ii
           z=z/ii;
           s=s+1;
       end
       if s>0
                str =[num2str(ii) '^' num2str(s) ];

                   string=strcat(string,str);
                   string= strcat(string,'*'); 
             % If z = 1, no more divisions are necessary, 
            % thus breaks the loop and quits
            if z == 1
                break
            end
        end

    end
    string=string(1:end-1);% remove last sign of multiplicaiton
fprintf('prime factorization is %s\n',string);

这里有几个例子

>> integer_factorization
enter your number : 30
prime factorization is 2^1*3^1*5^1

另一个

>> integer_factorization
enter your number : 35
prime factorization is 5^1*7^1

和最后一个

>> integer_factorization
enter your number : 100
prime factorization is 2^2*5^2

素数因子分解算法的格式化

5 个答案: