1

我编写了以下代码并投入了大量时间。但是它本质上是错误的，如果有人可以指导我提高效率，我将无限感激。

它目前不产生任何输出。

% A palindromic number reads the same both ways.
% The largest palindrome made from the product of
% two 2-digit numbers is 9009 = 91  99.

% Find the largest palindrome made from the
% product of two 3-digit numbers.

% 1) Find palindromes below 999x999 (product of two 3 digit #s)
% 2) For each palindrome found, find the greatest Factors.
% 3) The first palindrome to have two 3 digit factors is the
%    Solution



%============== Variables ===============================
%
% number = a product of 2 3 digit numbers, less than 100x100. The
% Palindrome we are looking for.
%
% n1, n2 = integers; possible factors of number.
%
% F1, F2 = two largest of factors of number. multiplied
% together they == number.
%
% finish = boolean variable, decides when the program terminates


% ================= Find Palindrome ================================

% The maximum product of two 3 digit numbers

number = 999*999;
finish = false;
count = 0;

while ( finish == false)

%
% Check to see if number is a Palindrome by comparing
% String versions of the number
%
% NOTE: comparing num2string vectors compares each element
% individually. ie, if both strings are identical, the output will be
% a vector of ONES whose size is equal to that of the two num2string
% vectors.
%
if ( mean(num2str( number ) == fliplr( num2str ( number ) ) ) == 1  )

    % fprintf(1, 'You have a palindrome %d', number);









    % Now find the greatest two factors of the discovered number ==========

    n1 = 100;
    n2 = 100; % temporary value to enter loop



    % While n2 has 3 digits in front of the decimal, continue
    % Searching for n1 and n2. In this loop, n1 increases by one
    % each iteration, and so n2 decreases by some amount. When n2
    % is no longer within the 3 digit range, we stop searching
    while( 1 + floor( log10( n2 ) ) == 3 )


        n2 = number/n1;



        % If n2 is EXACTLY a 3 digit integer,
        % n1 and n2 are 3 digit factors of Palindrome 'number'
        if( 1 + log10( n2 )  == 3 )

            finish = true;

            Fact1 = n1;
            Fact2 = n2;


        else
            % increment n1 so as to check for all possible
            % 3 digit factors ( n1 = [100,999] )
            n1 = n1 + 1;

        end

    end




    % if number = n1*n2 is not a palindrome, we must decrease one of the
    % Factors of number and restart the search
else

    count = count + 1;

    number = 999 * (999 - count);



end

end



fprintf(1, 'The largest factors of the palindrome %i \n', number )
fprintf(1, ' are %i and %i', Fact1, Fact2 )
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2 回答 2

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由于这是一个欧拉项目，我只会给出一些一般性的建议。

  1. 要比较字符串，请使用 strcmp，而不是您的方法（它使代码更简洁）

  2. 请参阅艾萨克的评论。添加floor 和检查数字是否为整数的条件（log10 不这样做）

  3. 即使你输入了那个 if 语句，你也永远不会真正退出它，因为 while 循环只是继续循环使用相同的两个数字。考虑通过修改你的while循环来终止while和那里。

  4. 虽然您的解决方案提供了一个结果，但它不是正确的，因为根据您的代码，数字始终是 999 的倍数，这很可能是不正确的。改变你的构建方式number。您必须至少添加另一行定义数字才能这样做。您的解决方案是 90909。正确的解决方案接近 100000（至少这是我发现的最高值）

于 2012-07-22T14:58:06.887 回答
1

条件：

if( 1 + log10( n2 )  == 3 )

仅当 时为真，并且当除法时n2 == 100​​看到 asn2仅是整数，您的循环可能永远不会完成。n1numberwhile

于 2012-07-21T22:40:54.787 回答