Question

我在stackoverflow Algorithm to find which number in a list sum up to a certain number

上找到了这个答案

我正在努力学习动态编程，并认为遵循逻辑是一项挑战。我尝试将S == 0改为其他值但这并没有产生预期的结果。

是否可以修改此代码，以便数字不必与总数完全相加，即允许小的差异来处理舍入问题。

def f(v, i, S, memo):
if i >= len(v): return 1 if S == 0 else 0
if (i, S) not in memo:  # <-- Check if value has not been calculated.
    count = f(v, i + 1, S, memo)
    count += f(v, i + 1, S - v[i], memo)
    memo[(i, S)] = count  # <-- Memoize calculated result.
return memo[(i, S)]     # <-- Return memoized value.

def g(v, S, memo):
subset = []
for i, x in enumerate(v):
# Check if there is still a solution if we include v[i]
    if f(v, i + 1, S - x, memo) > 0:
        subset.append(x)
        S -= x
return subset


v = [19783,21689,21169,11767,23207,12591]
sum = 56967
memo = dict()
if f(v, 0, sum, memo) == 0: print("There are no valid subsets.")
else: print(g(v, sum, memo))

返回[21169,23207,12591]

但是如果将21169更改为21168则返回“无有效子集”

v = [19783,21689,21168,11767,23207,12591]

子集和的模糊版本

0 个答案: