Matlab 项目4
我已经编写了下面的代码,并投入了大量的时间。但它本身就是有问题的,如果有人能善意地指导我提高效率,我将不胜感激 它目前不产生任何输出Matlab 项目4,matlab,Matlab,我已经编写了下面的代码,并投入了大量的时间。但它本身就是有问题的,如果有人能善意地指导我提高效率,我将不胜感激 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
% 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 )
条件:
if( 1 + log10( n2 ) == 3 )
只有当
n2==100
时才是真的,而当n1
除以number
时,n2才是整数,你的而循环可能永远不会结束。因为这是一个Euler项目,我只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(它有助于
清洁剂代码)
见艾萨克的评论。添加floor
和检查数字是否为整数的条件(log10不这样做)
即使您输入if语句,您也永远不会真正退出它,因为while循环只是继续使用相同的两个数字循环。考虑通过修改while循环来终止while,然后在中断的情况下。
虽然您的解决方案提供了一个结果,但它不是正确的结果,原因是根据您的代码,数字总是999的倍数,这很可能是错误的。更改构造编号的方式。为此,您必须至少添加一行定义数字。
您的解决方案是90909。正确的解决方案接近100000(至少是我发现的最高值)
Euler项目的要点是,你应该自己解决问题,而不是让别人替你解决。使用内置的探查器查看问题所在。在我的机器上短时间运行您的代码表示log10
正在消耗您的时间。想办法简化重复进行的计算。例如,如果您想知道一个编号n2
有三位数字,并且您知道它以三位数字开始并在减少,那么您真的需要记录该编号吗?不