将一个给定的数按递减顺序划分为一组整数的方法数
给定两个数字
和
。任务是找到用集合
表示 a 的方法数，使得
和这些数的和等于 a 。还有
(套件最大尺寸不能超过 m )。
示例:
输入 : a = 4，m = 4 输出:2–>({ 4 }、{3，1}) 注意 : {2，2}不是一个有效的集合，因为值不是按降序排列的
输入 : a = 7，m = 5 输出:5–>({ 7 }、{6，1}、{5，2}、{4，3}、{4，2，1})
方法:这个问题可以通过分治法使用递归方法来解决，该方法遵循以下条件:
- 如果 a 等于零，则找到了一个解。
- 如果 a > 0 且 m == 0，则该集合违反了条件，因为不能在该集合中添加更多的值。
- 如果已经对 a 、 m 和 prev(当前设置中包含的最后一个值)的给定值进行了计算，则返回该值。
- 从 i = a 到 0 开始循环，如果 i < prev ，如果我们在当前集合中包含 i ，则计算解的数量并返回。
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))
Output:
5
时间复杂度: O(a*log(a))
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