将给定数字降为一组整数的方式数量

给定两个数字$a$和$m$ 。任务是找到的其中一个可以由一组表示方式的数量$\{n_1, n_2, ...., n_c\}$这样$a >= n_1 > n_2 > ... > n_m > 0$这些数字的总和等于a 。还$1 <= c <= m$ (集合的最大大小不能超过m )。

例子：

Input : a = 4, m = 4
Output : 2 –> ({4}, {3, 1})
Note: {2, 2} is not a valid set as values are not in decreasing order

Input : a = 7, m = 5
Output : 5 –> ({7}, {6, 1}, {5, 2}, {4, 3}, {4, 2, 1})

方法：可使用以下条件的递归方法通过分而治之解决此问题：

• 如果a等于零，则找到一个解决方案。
• 如果a> 0且m == 0，则此集合违反条件，因为无法在集合中添加其他值。
• 如果已经针对给定的a ， m和prev(当前集中包含的最后一个值)进行了计算，则返回该值。
• 从i = a直到0开始循环，如果i < prev ，则在我们将i包含在当前集中并将其返回时，计算解决方案的数量。

下面是上述方法的实现：

# Python3 code to calculate the number of ways 
# in which a given number can be represented 
# as set of finite numbers
  
# Import function to initialize the dictionary
from collections import defaultdict
  
# Initialize dictionary which is used 
# to check if given solution is already 
# visited or not to avoid
# calculating it again
visited = defaultdict(lambda : False)
  
# Initialize dictionary which is used to
# store the number of ways in which solution
# can be obtained for given values
numWays = defaultdict(lambda : 0)
  
# This function returns the total number
# of sets which satisfy given criteria
# a --> number to be divided into sets
# m --> maximum possible size of the set
# x --> previously selected value
def countNumOfWays(a, m, prev):
      
    # number is divided properly and
    # hence solution is obtained
    if a == 0:
        return 1
      
    # Solution can't be obtained
    elif a > 0 and m == 0:
        return 0
      
    # Return the solution if it has
    # already been calculated
    elif visited[(a, m, prev)] == True:
        return numWays[(a, m, prev)]
      
    else:
        visited[(a, m, prev)] = True
  
        for i in range(a, -1, -1):
            # Continue only if current value is 
            # smaller compared to previous value
            if i < prev:
                numWays[(a,m,prev)] += countNumOfWays(a-i,m-1,i)
  
        return numWays[(a, m, prev)]
  
# Values of 'a' and 'm' for which
# solution is to be found
# MAX_CONST is extremely large value 
# used for first comparison in the function
a, m, MAX_CONST = 7, 5, 10**5
print(countNumOfWays(a, m, MAX_CONST))

输出：

5

时间复杂度： O(a * log(a))