#公雞5文錢(qián)一只 母雞3文錢(qián)一只 小雞1文錢(qián) 三只
money = 100
def buy_kun(gong:int,mu:int,xiao:int,c:int) -> None:
    if gong*5+mu*3+xiao > money or gong+mu+xiao*3>100:
        return
    if gong*5+mu*3+xiao == money  and gong+mu+xiao*3==100:
        print(f'買(mǎi){gong}只公雞，{mu}只母雞，{xiao*3}只小雞，剛好湊夠{money}文錢(qián)。')
        return
    if c<1 :
        buy_kun(gong+1,mu,xiao,0)
    if c<2:
        buy_kun(gong,mu+1,xiao,1)
    if c<3:
        buy_kun(gong,mu,xiao+1,2)
buy_kun(0,0,0,0)

# (公雞數(shù)，母雞數(shù)，小雞數(shù)，總錢(qián)數(shù)，總雞數(shù)),狀態(tài)轉(zhuǎn)移
import numpy as np  # 導(dǎo)入numpy庫(kù)
st = np.zeros((100, 100, 100))  # 新建三維數(shù)組，且初始值為0,記錄狀態(tài)是否被遍歷過(guò)，進(jìn)行剪枝操作
def resolve(a, b, c, d, e):
    if d == 100 and e == 100:
        print("%s只公雞，%s只母雞，%s只小雞" % (a, b, c))
    else:
        a1 = a + 1
        b1 = b + 1
        c1 = c + 3
        if d + 5 <= 100 and e + 1 <= 100 and st[a1][b][c] == 0:
            st[a1][b][c] = 1
            resolve(a1, b, c, d + 5, e + 1)
        if d + 3 <= 100 and e + 1 <= 100 and st[a][b1][c] == 0:
            st[a][b1][c] = 1
            resolve(a, b1, c, d + 3, e + 1)
        if d + 1 <= 100 and e + 3 <= 100 and st[a][b][c1] == 0:
            st[a][b][c1] = 1
            resolve(a, b, c1, d + 1, e + 3)
# (公雞數(shù)，母雞數(shù)，小雞數(shù)，總錢(qián)數(shù)，總雞數(shù))
resolve(0, 0, 0, 0, 0)

#公雞5文錢(qián)一只 母雞3文錢(qián)一只 小雞1文錢(qián) 三只
import numpy as np
st = np.zeros((100, 100, 100))
money = 100
def buy_kun(gong:int,mu:int,xiao:int) -> None:
    if st[gong][mu][xiao] == 1:
        return
    else:
        st[gong][mu][xiao] = 1
    if gong*5+mu*3+xiao > money or gong+mu+xiao*3>100:
        return
    if gong*5+mu*3+xiao == money  and gong+mu+xiao*3==100:
        print(f'買(mǎi){gong}只公雞，{mu}只母雞，{xiao*3}只小雞，剛好湊夠{money}文錢(qián)。')
        return
    buy_kun(gong+1,mu,xiao)
    buy_kun(gong,mu+1,xiao)
    buy_kun(gong,mu,xiao+1)
buy_kun(0,0,0)