[交流] Euler 工程 第三题：寻找600851475143的最大质因子已有7人参与

import timeit
from math import sqrt

def euler3():
    n = 600851475143

    primes = range(2, int(sqrt(n)))

    while n > 1:
        prime = primes[0]
        if n % prime == 0:
            print prime
            n = n / prime
        primes = [x for x in primes if x % prime != 0]


t = timeit.Timer("euler3.euler3()", "import euler3")
print t.timeit(1)

n = 600851475143;
tic;
primes = 2:round(sqrt(n));
while n > 1
    prime = primes(1);
    if mod(n, prime) == 0
        disp(num2str(prime));
        n = n / prime;
    end
    primes(find(mod(primes, prime)==0)) = [];
end
toc

#include
#include
#include

using namespace std;

int main(int argc, char** argv){

    double n = 600851475143.0;
    list primes;

    for(int i = 2; i < sqrt(n); i++)
        primes.push_back(i);

    while(n > 1.0){
        double prime = primes.front();
        if(fmod(n, prime) == 0){
            cout<             n /= prime;
        }

        list::iterator it;
        primes.erase(primes.begin());

        for(it = primes.begin(); it != primes.end(); it++){
            if(fmod(*it, prime) == 0)
                it = primes.erase(it);
        }
    }
}

clear;
primes = [2];
n = 600851475143;
tic;
while n > 1
    m = length(primes);
    prime = primes(m)+1:primes(m)+m;
    for i = 1:m
        prime(find(mod(prime, primes(i))==0)) = [];
    end
    for i = 1:length(prime)
        if mod(n, prime(i)) ==0
            fprintf('%d\n', prime(i));
            n = n / prime(i);
        end
    end
    primes = [primes prime];
end
toc

Originally posted by zdapeng at 2011-06-01 14:27:58:
#include<iostream.h>
#include<math.h>
using namespace std;
/*问题分析：
寻找最大质因数，只需要按从小到大的质数遍历
如果质素是该数的因数，则与之相除，直至不再是其因数
当该数变为1时 ...