Question

我编写了以下代码并花了很多时间。但是它本身存在一些本质上的错误，如果有人能指导我提高它的效率，我会非常感激。

它目前产生NO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 )

Answer 1

条件：

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

仅在n2 == 100时才会生效，并且在n2除n1时看到number只是一个整数，您的while循环可能永远不会光洁度。

Answer 2

由于这是一个欧拉项目，我只会给出一些一般性的建议。

要比较字符串，请使用strcmp，而不是您的方法（它可以实现清洁代码）
见Isaac的评论。添加floor 和条件以检查该数字是否为整数（log10不会这样做）
即使你输入if语句，你也永远不会退出它，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然后在那里休息。
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，这很可能是错误的。更改构建number的方式。您必须添加至少另一行定义数字才能这样做。你的解决方案是90909.正确的解决方案接近100000（至少是我发现的最高解决方案）

项目欧拉4

2 个答案: