c++ - 李白喝酒问题的算法

Question

“李白街上走，提壶去买酒，遇店加一倍，见花喝一斗”，途中，遇见5次店，见了10此花，壶中原有2斗酒，最后刚好喝完酒，要求最后遇见的是花，求可能的情况有多少种？希望大家分享一下思路，谢谢！
我的思路很混乱，觉得直接暴力枚举能解决，但是枚举所有的情况比较不现实，希望大家解答啊!

Answer 1

"""
这是今年蓝桥杯java中的一道填空题，当时我还以为做对了，写了33
结果我忽略了最明显的判断条件，5次店和10次花，少了下面的 check1 函数
简直为自己的智商捉鸡，加了之后就是14了，
最近学python，现用python实现这题，跟大家分享
"""

count=0;
def check(t):
    sum = 2;
    for ch in t:
        if ch=='1':
            sum=sum*2;
        if ch=='0':
            sum=sum-1;
    return sum;
def check1(t):
    c = 0;
    for ch in t:
        if(ch=='1'):
            c+=1;
    if c==5:
        return True;
    return False;
def f(t):
    if(len(t)==15):
        if check(t)==0 and t[14:15]=='0' and check1(t):
            global count;
            count+=1;
            print(" ".join(t));
        return;
    f(t+'1');
    f(t+'0');
f('');
print(count);

Answer 2

Simply give a non-recursive python code:

queue = [(2, 5, 10)]
total = 0
def iter():
    global queue
    while queue:
        i, queue = queue[0], queue[1:]
        yield i

for (left, dian, hua) in iter():
    if dian == 0:
        total += 1 if left == hua else 0
    else:
        if left > hua or left == 0 or hua ==0:
            continue
        queue.append((left*2, dian-1, hua))
        queue.append((left-1, dian, hua-1))

Answer 3

Violent exhaustion is easy to do, let’s not talk about it anymore
If the last time is a flower, the value of 2 is 1 after 14 transformations,
If it is 0 after seeing flowers for 10 times, the initial value is 2 buckets of wine, which means a total of 8 buckets of wine have been added
The minimum value for adding wine is 1, and the maximum value is 4 (11114),
Because adding wine is a doubling method, the difference between two adjacent numbers in the sequence cannot be greater than 2 times, that is, 11114 is excluded. The number of wine added may be in the range of 1-3. Also due to the reason of doubling, the number after 1 is only It can be 2 (but 3 is followed by 1)
Because the initial value is 2, the first time you add wine can only be 1 or 2 buckets
So just find a few series that meet the conditions