How can I solve this problem?

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David Lopez
David Lopez 2016-9-21
回答： RAMAKANT SHAKYA 2019-2-7
I am a bit lost with this Matlab problema that I have to do. Can anyone give me a hand? I would be very grateful!
" Write a function called circular_primes that finds the number of circular prime numbers smaller than n, where n is a positive integer scalar input argument. For example, the number, 197, is a circular prime because all rotations of its digits: 197, 971, and 719, are themselves prime. For instance, there are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97. It is important to emphasize that rotation means circular permutation not all possible permutations "
"2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder. Write a function called smallest_multiple that returns a uint64, the smallest positive number that is evenly divisible by all of the numbers from 1 to n where n is a positive integer scalar and is the only input argument of the function. If the result would be greater than what can be represented as a uint64, the function returns 0 "

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回答（2 个）

Srishti Saha
Srishti Saha 2018-3-11
This should work perfectly:
function mul = smallest_multiple(n)
facts = zeros(1,n); % store the exponents of various factors
for ii = 2:n
f = factor(ii); % get factors for current integer
for jj = 2:ii
k = sum(f == jj); % what's the exponent of this factor?
if k > facts(jj) % if it is greater than what we have so far
facts(jj) = k; % update to this new value
end
end
end
% Compute the result with one command.
% The 'native' option tells MATLAB to work in uint64
mul = prod(uint64((1:n).^facts),'native');
if mul == intmax('uint64')
mul = uint64(0);
end
end
RAMAKANT SHAKYA
RAMAKANT SHAKYA 2019-2-7
function out=circular_primes(no)
prim=primes(no);% find the all prime number till the given number
pr=0;
nos=[];
po=[];
for p=1:length(prim)
n=prim(p); % picking up each prime no one by one
n=num2str(n);% change into string for rotation of no
d=length(n); % length of string
if d>1 % take nos greater than 10 beacuase below 10 no need for rotation
for h=1:d
a=n(1);
for r=1:d % for rotation right to left
if r==d 5 % for the last element of string
n(d)=a;
else
n(r)=n(r+1); %shifting
end
end
s=str2num(n); % string to number
nos=[nos,s]; % store rotated elements in array
end
if nos(end)==no %if given no is also a circular prime we need smaller
break;
end
for gr=1:length(nos) % checking rotated nos are prime or not
p1=isprime(nos(gr));
po=[po,p1]; %storing logical result in array
end
if sum(po(:))==length(nos) %if all are prime the length and sum are must be equal
pr=pr+1;
out=pr;
else
out=pr;
end
po=[];
nos=[];
else
s=str2num(n); %numbers less than 10
f=isprime(s);
if f==1
pr=pr+1;
out=pr;
else
out=pr;
end
end
end
end

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