how to test for palindrome of the product of unknown digit numbers
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OLUBUKOLA ogunsola
2016 年 6 月 14 日
コメント済み: OLUBUKOLA ogunsola
2016 年 6 月 16 日
when testing out palindromic numbers for three digit numbers , we could create the products of the three digits this way:
for i=999:-1:100
for j=999:-1:100
answer = i*j;
what if we do not know the number of digit and the number of digit will be the input to the code , how can i adjust the 'for loop' to cater for the unknown number of digits?
say input could be 2 digits or 4 digits or 5 digits and the code will create the products
thanks
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採用された回答
Walter Roberson
2016 年 6 月 14 日
for j = 10^(number_of_digits) - 1 : -1 : 10^(number_of_digits-1)
5 件のコメント
Walter Roberson
2016 年 6 月 15 日
I do not see anywhere in your code where your store anything into a matrix.
I tested my code from number_of_digits = 1 to number_of_digits = 4 and it works fine.
その他の回答 (1 件)
Image Analyst
2016 年 6 月 15 日
Try this:
% Code with a palindrome.
theNumber = 123848321; % A palindrome
str = sprintf('%d', theNumber)
if all(str == fliplr(str))
message = sprintf('%d is a palindrome', theNumber);
uiwait(helpdlg(message));
else
message = sprintf('%d is NOT a palindrome', theNumber);
uiwait(helpdlg(message));
end
% Code with a non-palindrome.
theNumber = 163848321; % NOT a palindrome
str = sprintf('%d', theNumber)
if all(str == fliplr(str))
message = sprintf('%d is a palindrome', theNumber);
uiwait(helpdlg(message));
else
message = sprintf('%d is NOT a palindrome', theNumber);
uiwait(helpdlg(message));
end
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