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[½»Á÷] Project Euler 50 Å·À­¹¤³Ì 50 Ìâ ÒÑÓÐ6È˲ÎÓë

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                41 = 2 + 3 + 5 + 7 + 11 + 13
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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?

ºÇºÇ¡­¡­ÏÐ×ÅûÊ£¬ÎÒÒ²ÌùÒ»¸ö£¬¿´¿´´ó¼Ò»¹ÓÐûÓÐÈÈÇé¡­¡­

[ Last edited by ΢³¾¡¢ÃÎÏë on 2012-2-17 at 22:28 ]
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CODE:
#!/usr/bin/env python

def generate_primes(n):
    isprimes=[True]*n
    for i in range(2,n):
        if isprimes[i]:
            for j in range(2*i,n,i):
                isprimes[j] = False
    primes=[i for i in range(3,n,2) if isprimes[i]]
    primes.insert(0,2)
    return primes


def euler50(num):
    p=generate_primes(num)
    max_n=1
    while sum(p[:max_n])         max_n+=1
   
    for i in range(max_n,2,-1):
        for j in range(len(p)-i):
            tmp=sum(p[j:j+i])
            if tmp>num:break
            if tmp in p:
                return tmp
        
if __name__ == "__main__":
    print euler50(1000000)

[ Last edited by wangww2011 on 2012-2-20 at 03:45 ]
3Â¥2012-02-20 03:43:42
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matlab code
CODE:
% 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

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