我已经写了下面的代码,并投入了大量的时间。但是有一些本质上的错误,如果有人能够指导我提高效率,我会感激不尽。项目欧拉4
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)
与项目欧拉的一点是,你应该自己弄清楚,不要让别人代劳为你。 – Guffa 2012-07-21 20:55:55
使用内置分析器查看挂机位置。在我的机器上短时间运行你的代码说''log10'正在浪费你的时间。想想如何简化重复发生的计算。例如,如果你想知道一个数字'n2'有三位数字,并且你知道它以三位数字开始并且正在减少,你是否真的需要记录那个数字的日志?没有。 – zroth 2012-07-21 21:02:00