前n个自然数立方和的Python程序-yiteyi-C++库

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：

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：

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：

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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