改进我的euler p12项目代码。在Python中

Question

我在做项目欧拉问题12，并提出了以下代码，在对低数字进行测试时效果良好：

def highlyDivisibleTriangularNumber(n):
    """
    triangular num formula: x(x-1)/2
     e.g 5th triang num: x5 = 5(5+1)/2 = 15

    :param n:
    """
    divisors = [x for x in xrange(1, 31)]
    count = 0
    for i in xrange(1, n):
        triangNum  = (i* (i+1)/2)         # create triangular number based on formula: x(x-1)/2
        for div in divisors:
            if triangNum % div == 0:      # if triangular number modulo div equals 0 the add it to highestDivisors
                count +=1
                if count == 6:              # if count  is equal to 6 then  we have found the  required number so return it.
                            return triangNum
            else:
                if div == 30:                      # else if we have have reached the  last divisor, reset count to zero
                    count = 0

print(highlyDivisibleTriangularNumber(100))

我将非常感谢您对如何提高效率的一些建议，我在哪里出错，以及在设置除数上限方面。 我确信从数学的角度来看，无疑有许多方法可以改进代码。 谢谢。在

三角形数序列是由自然数相加生成的。所以第七个三角形是1+2+3+4+5+6+7=28。前十个术语是：

1，3，6，10，15，21，28，36，45，55。。。在

让我们列出前七个三角形数的因子：

1比1 3： 1，3 6： 1,2,3,6 10： 1,2,5,10 15： 1,3,5,15 21:1,3,7,21 28:1,2,4,7,14,28 我们可以看到28是第一个有5个以上除数的三角形数。在

第一个有超过500个除数的三角形数的值是多少？在

Answer 1

# a list pre calculated of  prime below 1000, I just don't want to write the
# cal prime code here, it is not the bottleneck
prime_list= [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997, 1009, 1013, 1019, 1021, 1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069, 1087, 1091, 1093, 1097, 1103, 1109, 1117, 1123, 1129, 1151, 1153, 1163, 1171, 1181, 1187, 1193, 1201, 1213, 1217, 1223, 1229, 1231, 1237, 1249, 1259, 1277, 1279, 1283, 1289, 1291, 1297, 1301, 1303, 1307, 1319, 1321, 1327, 1361, 1367, 1373, 1381, 1399, 1409, 1423, 1427, 1429, 1433, 1439, 1447, 1451, 1453, 1459, 1471, 1481, 1483, 1487, 1489, 1493, 1499, 1511, 1523, 1531, 1543, 1549, 1553, 1559, 1567, 1571, 1579, 1583, 1597, 1601, 1607, 1609, 1613, 1619, 1621, 1627, 1637, 1657, 1663, 1667, 1669, 1693, 1697, 1699, 1709, 1721, 1723, 1733, 1741, 1747, 1753, 1759, 1777, 1783, 1787, 1789, 1801, 1811, 1823, 1831, 1847, 1861, 1867, 1871, 1873, 1877, 1879, 1889, 1901, 1907, 1913, 1931, 1933, 1949, 1951, 1973, 1979, 1987, 1993, 1997, 1999, 2003, 2011, 2017, 2027, 2029, 2039, 2053, 2063, 2069, 2081, 2083, 2087, 2089, 2099, 2111, 2113, 2129, 2131, 2137, 2141, 2143, 2153, 2161, 2179, 2203, 2207, 2213, 2221, 2237, 2239, 2243, 2251, 2267, 2269, 2273, 2281, 2287, 2293, 2297, 2309, 2311, 2333, 2339, 2341, 2347, 2351, 2357, 2371, 2377, 2381, 2383, 2389, 2393, 2399, 2411, 2417, 2423, 2437, 2441, 2447, 2459, 2467, 2473, 2477, 2503, 2521, 2531, 2539, 2543, 2549, 2551, 2557, 2579, 2591, 2593, 2609, 2617, 2621, 2633, 2647, 2657, 2659, 2663, 2671, 2677, 2683, 2687, 2689, 2693, 2699, 2707, 2711, 2713, 2719, 2729, 2731, 2741, 2749, 2753, 2767, 2777, 2789, 2791, 2797, 2801, 2803, 2819, 2833, 2837, 2843, 2851, 2857, 2861, 2879, 2887, 2897, 2903, 2909, 2917, 2927, 2939, 2953, 2957, 2963, 2969, 2971, 2999, 3001, 3011, 3019, 3023, 3037, 3041, 3049, 3061, 3067, 3079, 3083, 3089, 3109, 3119, 3121, 3137, 3163, 3167, 3169, 3181, 3187, 3191, 3203, 3209, 3217, 3221, 3229, 3251, 3253, 3257, 3259, 3271, 3299, 3301, 3307, 3313, 3319, 3323, 3329, 3331, 3343, 3347, 3359, 3361, 3371, 3373, 3389, 3391, 3407, 3413, 3433, 3449, 3457, 3461, 3463, 3467, 3469, 3491, 3499, 3511, 3517, 3527, 3529, 3533, 3539, 3541, 3547, 3557, 3559, 3571, 3581, 3583, 3593, 3607, 3613, 3617, 3623, 3631, 3637, 3643, 3659, 3671, 3673, 3677, 3691, 3697, 3701, 3709, 3719, 3727, 3733, 3739, 3761, 3767, 3769, 3779, 3793, 3797, 3803, 3821, 3823, 3833, 3847, 3851, 3853, 3863, 3877, 3881, 3889, 3907, 3911, 3917, 3919, 3923, 3929, 3931, 3943, 3947, 3967, 3989, 4001, 4003, 4007, 4013, 4019, 4021, 4027, 4049, 4051, 4057, 4073, 4079, 4091, 4093, 4099, 4111, 4127, 4129, 4133, 4139, 4153, 4157, 4159, 4177, 4201, 4211, 4217, 4219, 4229, 4231, 4241, 4243, 4253, 4259, 4261, 4271, 4273, 4283, 4289, 4297, 4327, 4337, 4339, 4349, 4357, 4363, 4373, 4391, 4397, 4409, 4421, 4423, 4441, 4447, 4451, 4457, 4463, 4481, 4483, 4493, 4507, 4513, 4517, 4519, 4523, 4547, 4549, 4561, 4567, 4583, 4591, 4597, 4603, 4621, 4637, 4639, 4643, 4649, 4651, 4657, 4663, 4673, 4679, 4691, 4703, 4721, 4723, 4729, 4733, 4751, 4759, 4783, 4787, 4789, 4793, 4799, 4801, 4813, 4817, 4831, 4861, 4871, 4877, 4889, 4903, 4909, 4919, 4931, 4933, 4937, 4943, 4951, 4957, 4967, 4969, 4973, 4987, 4993, 4999, 5003, 5009, 5011, 5021, 5023, 5039, 5051, 5059, 5077, 5081, 5087, 5099, 5101, 5107, 5113, 5119, 5147, 5153, 5167, 5171, 5179, 5189, 5197, 5209, 5227, 5231, 5233, 5237, 5261, 5273, 5279, 5281, 5297, 5303, 5309, 5323, 5333, 5347, 5351, 5381, 5387, 5393, 5399, 5407, 5413, 5417, 5419, 5431, 5437, 5441, 5443, 5449, 5471, 5477, 5479, 5483, 5501, 5503, 5507, 5519, 5521, 5527, 5531, 5557, 5563, 5569, 5573, 5581, 5591, 5623, 5639, 5641, 5647, 5651, 5653, 5657, 5659, 5669, 5683, 5689, 5693, 5701, 5711, 5717, 5737, 5741, 5743, 5749, 5779, 5783, 5791, 5801, 5807, 5813, 5821, 5827, 5839, 5843, 5849, 5851, 5857, 5861, 5867, 5869, 5879, 5881, 5897, 5903, 5923, 5927, 5939, 5953, 5981, 5987, 6007, 6011, 6029, 6037, 6043, 6047, 6053, 6067, 6073, 6079, 6089, 6091, 6101, 6113, 6121, 6131, 6133, 6143, 6151, 6163, 6173, 6197, 6199, 6203, 6211, 6217, 6221, 6229, 6247, 6257, 6263, 6269, 6271, 6277, 6287, 6299, 6301, 6311, 6317, 6323, 6329, 6337, 6343, 6353, 6359, 6361, 6367, 6373, 6379, 6389, 6397, 6421, 6427, 6449, 6451, 6469, 6473, 6481, 6491, 6521, 6529, 6547, 6551, 6553, 6563, 6569, 6571, 6577, 6581, 6599, 6607, 6619, 6637, 6653, 6659, 6661, 6673, 6679, 6689, 6691, 6701, 6703, 6709, 6719, 6733, 6737, 6761, 6763, 6779, 6781, 6791, 6793, 6803, 6823, 6827, 6829, 6833, 6841, 6857, 6863, 6869, 6871, 6883, 6899, 6907, 6911, 6917, 6947, 6949, 6959, 6961, 6967, 6971, 6977, 6983, 6991, 6997, 7001, 7013, 7019, 7027, 7039, 7043, 7057, 7069, 7079, 7103, 7109, 7121, 7127, 7129, 7151, 7159, 7177, 7187, 7193, 7207, 7211, 7213, 7219, 7229, 7237, 7243, 7247, 7253, 7283, 7297, 7307, 7309, 7321, 7331, 7333, 7349, 7351, 7369, 7393, 7411, 7417, 7433, 7451, 7457, 7459, 7477, 7481, 7487, 7489, 7499, 7507, 7517, 7523, 7529, 7537, 7541, 7547, 7549, 7559, 7561, 7573, 7577, 7583, 7589, 7591, 7603, 7607, 7621, 7639, 7643, 7649, 7669, 7673, 7681, 7687, 7691, 7699, 7703, 7717, 7723, 7727, 7741, 7753, 7757, 7759, 7789, 7793, 7817, 7823, 7829, 7841, 7853, 7867, 7873, 7877, 7879, 7883, 7901, 7907, 7919]
n = 1
while True:
    triangle  = n*(n+1)/2
    _triangle = triangle
    prime_factors = []
    for prime in prime_list:
        counter = 0
        while _triangle % prime == 0:
            _triangle /= prime
            counter +=1
        prime_factors.append(counter+1)
    # If triangle = p1^q1*p2^q2*p3^3 ... p1,p2 ... are primes
    # the number of divisors is (q1+1)*(q2+1)*(q3+1)...
    # The math is really simple here:
    # any divisor of triangle can be written as : p1^r1*p2^r2*p3^r3 (r1's range is 0 to q1)
    divisor_num = reduce(lambda a,b : a*b, prime_factors)
    if divisor_num > 500 :
        print triangle
        break
    else:
        n += 1

大约需要5-10秒，不是很快，但可以接受。在

Answer 2

下面所示的代码避免了前面介绍的代码中存在的几个问题。在

原始post中的一个问题是通过测试所有小于正在处理的t的数字来计算除数。存在各种可能的加速：不要测试(t+2)/2和t-1之间的数字，因为它们都不能除以t。这可以扩展（例如，(t+3)/3和{}之间的数字不需要测试），但回报率会迅速减少。更重要的加速是使用divisor function的乘法形式；例如，如果一个数t等于 pᵃ·qᵇ·rᶜ带p, q, r素数，然后 σ₀(t) = τ(t) = (a+1)·(b+1)·(c+1)。（除数计数函数σ₀（sigma zero）通常用τ（tau）表示，而不是σ₀。）

另一个答案中的一个问题是在每个步骤中(i*(i+1)/2)或{}进行因式分解。在每个步骤中只考虑i+1或{}，并记住上一步中的因子会更快。这样可以将执行时间缩短至少2倍。在

下面的代码通过τ(k)·τ(k+1)·e/(e+1))计算τ(k·(k+1)/2)，其中e表示乘积τ(k)·τ(k+1)中2的指数。也就是说，因为k的素因子不能是k+1的素因子，反之亦然，τ(k·(k+1)) = τ(k)·τ(k+1)；并且{}比{}小。在

当findHiTau(120)调用时，附加的程序在我的旧计算机上运行大约0.14秒。findHiTau的参数是程序oddprimes表中素数大小的上限，是被测试的k的最高值的平方根；因此将处理大约k⁴/2的三角形数。在

# Returns divisor count for integer v and power of 2 in v
def tau(v):
    p2 = 0
    while v%2 == 0:
        v, p2 = v/2, p2+1
    t = p2+1
    for p in oddprimes:
        if v%p == 0:
            e = 1
            while v%p == 0:
                v, e = v/p, e+1
            t *= e
        if p*p > v: break
    if v>1:                     # Rest of v is a prime
        t *= 2
    return (t, p2)

def findHiTau(primeLim):   # Show record-high values of tau(k*(k+1)/2) up to k=m
    # Make array of small odd primes (expression is ok at least to 100K)
    oddprimes = [3,5,7,11,13]+[x for x in range(17,primeLim,2) if 1==pow(2,x-1,x)==pow(3,x-1,x)==pow(5,x-1,x)==pow(7,x-1,x)==pow(11,x-1,x)==pow(13,x-1,x)]
    taub, e2, hitau = 2, 1, 0       # e2 = exponent of 2
    # In loop, taua = tau(k-1); taub = tau(k); e2 = power of 2 in (k-1)*k
    for k in range(3,primeLim*primeLim):
        taua = taub
        (taub, f2) = tau(k)
        if f2:  e2 = f2         # If f is even, update e2
        ttau = taua*taub*e2/(e2+1)
        if ttau > hitau:
            hitau = ttau
            print '{:3} {:3}  {:3}'.format(k-1, k, ttau)