Fibonacci兔在任意月份死亡

Question

因此，我看到了一些解决这个问题或类似问题的方法，但我真的想知道为什么我的解决方案不起作用。它比我找到的许多解决方案更容易阅读，所以我很乐意让它工作！

从1只兔子开始，2个月后开始繁殖。跑了n个月，兔子活了m个月就死了。输入的'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问题中的答案中复制出来的，以帮助将其从“未回答”的问题列表中剔除。

这里的两个键将树画成一个大的数字，并确保包含了第一代和第二代死亡病例的基本情况检查(在这两种情况下都是-1，然后取决于输入)。

因此，有3起潜在案件。规则的fib序列，当我们不需要考虑死亡，第一代和第二代死亡初始化我们的最终序列与重复关系Fn-2 + Fn-1 - Fn-(monthsAlive+1)

我确信有一种方法可以合并1或2个这样的检查，并使算法更高效，但到目前为止，它立即正确地解决了一个大型测试用例(90，17)。所以我很高兴。

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

复发关系2例。考虑到n是序列运行的月份数，m是一对要生存的月数：

1)如果序列中的索引(基于零的)小于m

正规Fibonacci (当前项=前一项+它之前的项)。

2)如果索引大于或等于m

当前项=和(m - 1)以前的项(忽略前面的项)。

下面是一个名为A和m =5的序列的示例： A5 = A0 + A1 + A2 + A3 (4个术语，即m - 1，忽略前面的一个) 。 如果m = 3，那么： A3 = A0 + A1 (仅2项，m - 1)

。

下面的代码(在Python中)的偏移量为2，这是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