前n个自然数立方和的Python程序

打印系列1的总和 3. + 2 3. + 3 3. + 4 3. + …….+ N 3. 直到第n学期。

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例如:

Input : n = 5
Output : 225
13 + 23 + 33 + 43 + 53 = 225

Input : n = 7
Output : 784
13 + 23 + 33 + 43 + 53 + 
63 + 73 = 784

Python3

# Simple Python program to find sum of series
# with cubes of first n natural numbers
# Returns the sum of series
def sumOfSeries(n):
sum = 0
for i in range ( 1 , n + 1 ):
sum + = i * i * i
return sum
# Driver Function
n = 5
print (sumOfSeries(n))
# Code Contributed by Mohit Gupta_OMG <(0_o)>


输出:

225

时间复杂度:O(n) 一 有效解决方案 就是使用直接的数学公式 (n(n+1)/2)^2

For n = 5 sum by formula is
       (5*(5 + 1 ) / 2)) ^ 2
          = (5*6/2) ^ 2
          = (15) ^ 2
          = 225

For n = 7, sum by formula is
       (7*(7 + 1 ) / 2)) ^ 2
          = (7*8/2) ^ 2
          = (28) ^ 2
          = 784

Python3

# A formula based Python program to find sum
# of series with cubes of first n natural
# numbers
# Returns the sum of series
def sumOfSeries(n):
x = (n * (n + 1 ) / 2 )
return ( int )(x * x)
# Driver Function
n = 5
print (sumOfSeries(n))
# Code Contributed by Mohit Gupta_OMG <(0_o)>


输出:

225

时间复杂度:O(1) 这个公式是如何工作的? 我们可以用数学归纳法来证明这个公式。我们可以很容易地看到,这个公式适用于n=1和n=2。对于n=k-1,这是真的。

Let the formula be true for n = k-1.
Sum of first (k-1) natural numbers = 
            [((k - 1) * k)/2]2

Sum of first k natural numbers = 
          = Sum of (k-1) numbers + k3
          = [((k - 1) * k)/2]2 + k3
          = [k2(k2 - 2k + 1) + 4k3]/4
          = [k4 + 2k3 + k2]/4
          = k2(k2 + 2k + 1)/4
          = [k*(k+1)/2]2

即使结果不超过整数限制,上述程序也会导致溢出。 喜欢 以前的职位 ,我们可以通过先除法在一定程度上避免溢出。

Python3

# Efficient Python program to find sum of cubes
# of first n natural numbers that avoids
# overflow if result is going to be withing
# limits.
# Returns the sum of series
def sumOfSeries(n):
x = 0
if n % 2 = = 0 :
x = (n / 2 ) * (n + 1 )
else :
x = ((n + 1 ) / 2 ) * n
return ( int )(x * x)
# Driver Function
n = 5
print (sumOfSeries(n))
# Code Contributed by Mohit Gupta_OMG <(0_o)>


输出:

225

请参阅完整的文章 前n个自然数的立方和程序 更多细节!

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