Highly divisible triangular number Problem 12 The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ... Let us list the factors of the first seven triangle numbers: 1: 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 We can see that 28 is the first triangle number to have over five divisors. What is the value of the first triangle number to have over five hundred divisors?
def divisor(num,j):
count1 = 1
count2 = 1
if num%2 == 0:
lst1 = [num//2]
lst2 = [num+1]
else:
lst1 = [(num+1)//2]
lst2 = [num]
for i in range(2,lst1[0]):
if lst1[0] % i ==0:
count1 += 1
# print(i)
lst1.append(i)
for i in range(2,lst2[0]):
if lst2[0] % i ==0:
count2 += 1
lst2.append(i)
count = count1 + count2 + count1*count2
return count n = 1
import time
print(time.time())
while True:
j = n*(n+1)//2
if str(j)[-1] != '':
#print(n)
n += 1
continue
count = divisor(n,j)
# print(j,n,count)
n += 1
if count >= 500:
print(j,n,count)
break print(time.time()) 1462342495.953848
76576500 12376 575
1462342499.651143
约4S
对求因子的算法进行一点改进,能提高一半效率。
for i in range(2,lst1[0]//2+1): for i in range(2,lst2[0]//2+1):
>>>
1462414510.288464
76576500 12376 575
1462414512.207313
再次对求因子进行改进,效率提高10倍!!
for i in range(2,int(sqrt(lst1[0])+1)):
if lst1[0] % i ==0:
count1 += 2
1462415389.778016
76576500 12376 575
1462415389.91842
>>>