Question

所以，我已经看到了针对这个问题或类似问题的一些解决方案，但我真的想知道为什么我的工作不起作用。它比我找到的很多解决方案更容易阅读，所以我很乐意让它发挥作用！

从1对兔开始，2个月后开始繁殖。跑了n个月，兔子在生活了几个月后就死了。输入'6 3'应该返回4，但它返回3.

#run for n months, rabbits die after m months.
n, m = input("Enter months to run, and how many months rabbits live, separated by a space ").split() 
n, m = int(n), int(m)
generations = [1, 1, 2] #Seed the sequence with the 1 pair, then in their reproductive month.
def fib(i, j):
    count = 3 #we start at the 3rd generation.
    while (count < i):
        if (count < j):
            generations.append(generations[count-2] + generations[count-1]) #recurrence relation before rabbits start dying
        else:                                                               #is just the fib seq (Fn = Fn-2 + Fn-1)
            generations.append((generations[count-2] + generations[count-1]) - generations[(count-j)])  #Our recurrence relation when rabbits die every month
        count += 1                                                          #is (Fn = Fn-2 + Fn-1 - Fn-j)
    return (generations[count-1])


print (fib(n, m))
print ("Here's how the total population looks by generation: \n" + str(generations))

谢谢=]

Answer 1

这是从SpaceCadets问题的答案中复制的，以帮助将其从“未答复”的问题列表中删除。

这里的两个关键是将树大量抽出来，并确保包含第一代和第二代死亡的基础案例检查（在两种情况下都是-1，然后它取决于输入）。

所以3个潜在的案例。我们不需要考虑死亡时的常规纤维序列，第一代和第二代死亡以初始化我们的最终序列与复发关系Fn-2 + Fn-1 - Fn-（monthsAlive + 1）

我确信有一种方法可以合并这些检查中的1个或2个并使算法更有效，但到目前为止它立即正确地解决了大型测试用例（90,17）。所以我很高兴。

获得的经验教训：使用整个白板。

#run for n months, rabbits die after m months.
n, m = input("Enter months to run, and how many months rabbits live, separated by a space ").split() 
n, m = int(n), int(m)
generations = [1, 1] #Seed the sequence with the 1 pair, then in their reproductive month.
def fib(i, j):
    count = 2
    while (count < i):
        if (count < j):
            generations.append(generations[-2] + generations[-1]) #recurrence relation before rabbits start dying (simply fib seq Fn = Fn-2 + Fn-1)
        elif (count == j or count == j+1):
            print ("in base cases for newborns (1st+2nd gen. deaths)") #Base cases for subtracting rabbit deaths (1 death in first 2 death gens)
            generations.append((generations[-2] + generations[-1]) - 1)#Fn = Fn-2 + Fn-1 - 1
        else:
            generations.append((generations[-2] + generations[-1]) - (generations[-(j+1)])) #Our recurrence relation here is Fn-2 + Fn-1 - Fn-(j+1)
        count += 1
    return (generations[-1])


print (fib(n, m))
print ("Here's how the total population looks by generation: \n" + str(generations))

Answer 2

使用递归。

public static int fibRec(int months, int dieAfter) {
    if(months <= 0) return 0;
    if(months == 1) return 1;
    if(months <= dieAfter)
        return fibRec(months-1, dieAfter) + fibRec(months-2, dieAfter);
    else if (months == dieAfter+1)
        return fibRec(months-1, dieAfter) + fibRec(months-2, dieAfter) - 1;
    else
        return fibRec(months-1, dieAfter) + fibRec(months-2, dieAfter) 
                 - fibRec(months-(dieAfter+1), dieAfter);
}

Answer 3

这里有2例复发关系。考虑 n 是序列运行的月数， m 是一对将要生活的月数：

1）如果序列中的索引（从零开始）小于 m ：
普通斐波纳契（当前术语=上一个术语+之前的术语）。

2）如果指数大于或等于 m ：
当前术语= （ m - 1 ）之前的术语总和（忽略之前的术语）。

以下是一个名为 A 且 m = 5 的序列的示例：
  A5 = A0 + A1 + A2 + A3（4个术语，即 m - 1 ，忽略前一个）     。
  如果 m = 3 那么：
  A3 = A0 + A1（仅2个术语， m - 1 ）

。

下面的代码（在Python中）的偏移量为2，因为序列开头的初始值为[1,1]。

def mortal_rabbits(n, m):
    sequence = [1, 1]

    for i in range(n - 2):
        new_num = 0
        if i + 2 < m:
            #Normal fibonacci - No deaths yet
            new_num = sequence[i] + sequence[i + 1]
        else:
            #Different reoccurence relation - Accounting for death
            for j in range(m - 1):
                new_num += sequence[i - j]

        sequence.append(new_num)

    return sequence

Answer 4

忽略并发症，一代中兔子对数的等式是

$R(n,m) = R(n-1) + R(n-2) - R(n - (m + 1))$

如果我们使用列表，则会出现问题，因为在n == m时，我们需要位于-1的值，这显然是超出范围的。

def rabbit_pairs(n, m):
    sequence = list()
    for i in range(n):
        if i < 2:
            # Normal Fibonacci initialization
            total = 1
            sequence.append(total)
        elif (i < m) or (m == 0):
            # Normal Fibonacci calculation
            total = sequence[i - 1] + sequence[i - 2]
            sequence.append(total)
        elif i == m:
            # Now we need R(n - (m + 1)), but i - (m + 1) < 0, so we have to
            # provide the missing value
            total = sequence[i - 1] + sequence[i - 2] - 1
            sequence.append(total)
        else:
            # i - (m + 1) >= 0, so we can get the value from the sequence
            total = sequence[i - 1] + sequence[i - 2] - sequence[i - (m + 1)]
            sequence.append(total)
    return total

Answer 5

天真地模拟任务中的行为的代码，我猜可以更快地使用touples

def fibm(n,m):
    seq = [[0,0],[0,1],[1,0],[1,1]]
    
    #(old,new)
    for i in range(4,n+1):
        
        new = seq[i-1][0]#new = previous old
        old = seq[i-1][0] + seq[i-1][1] #old = prev old + prev new
        
        if i>m:
            
            old = old - seq[i-m][1] #new from m ago die off
            
        seq.append([old,new])
    return(seq)       
                   
n,m = 95,16

print(sum(fibm(n,m)[-1]))

Fibonacci兔子在任意数月之后死亡

5 个答案: