Math 带约束的n元方程解的个数
我想计算方程的解的数量,但我无法获得任何线索。方程式为: 我所能做的就是Math 带约束的n元方程解的个数,math,multinomial,Math,Multinomial,我想计算方程的解的数量,但我无法获得任何线索。方程式为: 我所能做的就是 但我不知道如何处理这个问题。我会尝试使用 下面是一些伪代码,让您开始: Procedure num_solutions(n, k, m): # Initialize memoization cache: if this function has been called for the first time: initialize memo_cache with (n+1)*(k+1)*(m+1) ele
但我不知道如何处理这个问题。我会尝试使用 下面是一些伪代码,让您开始:
Procedure num_solutions(n, k, m):
# Initialize memoization cache:
if this function has been called for the first time:
initialize memo_cache with (n+1)*(k+1)*(m+1) elements, all set to -1
# Return cached solution if available
if memo_cache[n][k][m] is not -1:
return memo_cache[n][k][m]
# Edge case:
if m is equal to 1:
# Solution only exists if 1 <= m <= k
if n >= 1 and n <= k, set memo_cache[n][k][m] to 1 and return 1
otherwise set memo_cache[n][k][m] to 0 and return 0
# Degenerate case: No solution possible if n<m or n>k*m
if n < m or n > k * m:
set memo_cache[n][k][m] to 0 and return 0
# Call recursively for a solution with m-1 elements
set sum to 0
for all i in range 1..k:
sum = sum + num_solutions(n - i, k, m - 1)
set memo_cache[n][k][m] to sum and return sum
程序编号(n,k,m):
#初始化备忘录缓存:
如果此函数是第一次调用:
使用(n+1)*(k+1)*(m+1)个元素初始化memo_缓存,所有元素都设置为-1
#返回缓存的解决方案(如果可用)
如果memo_cache[n][k][m]不是-1:
返回备忘录\u缓存[n][k][m]
#边缘情况:
如果m等于1:
#只有当1应用几何和,然后使用分母的二项式定理和二项式级数时,解才存在。