欧拉计划(python) problem 65

时间：2015-01-31 16:27:32 阅读：178 评论：0 收藏：0 [点我收藏+]

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Convergents of e

Problem 65

The square root of 2 can be written as an infinite continued fraction.

√2 = 1 +	1
	2 +	1
		2 +	1
			2 +	1
				2 + ...

The infinite continued fraction can be written, √2 = [1;(2)], (2) indicates that 2 repeats ad infinitum. In a similar way, √23 = [4;(1,3,1,8)].

It turns out that the sequence of partial values of continued fractions for square roots provide the best rational approximations. Let us consider the convergents for √2.

1 +	1	= 3/2
	2

1 +	1		= 7/5
	2 +	1
		2

1 +	1			= 17/12
	2 +	1
		2 +	1
			2

1 +	1				= 41/29
	2 +	1
		2 +	1
			2 +	1
				2

Hence the sequence of the first ten convergents for √2 are:

1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, ...

What is most surprising is that the important mathematical constant,
e = [2; 1,2,1, 1,4,1, 1,6,1 , ... , 1,2k,1, ...].

The first ten terms in the sequence of convergents for e are:

2, 3, 8/3, 11/4, 19/7, 87/32, 106/39, 193/71, 1264/465, 1457/536, ...

The sum of digits in the numerator of the 10^th convergent is 1+4+5+7=17.

Find the sum of digits in the numerator of the 100^th convergent of the continued fraction for e.

python code :

def func(result,i):
t=i*result[1]+result[0]
result[0]=result[1]
result[1]=t
return result

def func1(x):
if(x%3)==0:
k=2*(x/3)
else:
k=1
return k

n=100
result=[1,func1(n)]
for i in range(n-1,1,-1):
result=func(result,func1(i))

result[0]+=2*result[1]
temp=str(int(result[0]))
result=0
k=len(temp)
for i in range(0,k):
result+=int(temp[i])
print(result)

time : <1s

欧拉计划(python) problem 65

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原文地址：http://blog.csdn.net/zhangzhengyi03539/article/details/43340391

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