[交流] Project Euler 50 欧拉工程 50 题已有6人参与

The prime 41, can be written as the sum of six consecutive primes:

41 = 2 + 3 + 5 + 7 + 11 + 13
This is the longest sum of consecutive primes that adds to a prime below one-hundred.

The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms, and is equal to 953.

Which prime, below one-million, can be written as the sum of the most consecutive primes?

% Elapsed time is 0.083633 seconds.
% ans =
%       997651
function result = euler50()
tic;
result = 0;
maxLen = 0;

ap = primes(1e6);

sumlist = cumsum(ap);
sumlist(sumlist>=1e6) = [];

maxLen = 21;
for i=1:length(sumlist)
    for j=i+maxLen:length(sumlist)
        temp = sumlist(j)-sumlist(i);
        if isprime(temp) && (j-i)>maxLen
            maxLen = j-i;
            result = temp;
        end
    end
end
toc;
end

4楼: Originally posted by lurencyj at 2012-02-20 17:26:22:
听说python效率比C++差不多慢个100倍。。。哈哈