题解 | #百钱买百鸡问题#

方法1 :python 枚举遍历法

def GetResult():
    factor = [5,3,1/3]
    L = []
    for gj in range(100//factor[0]):
        mj_num = (100 - factor[0]*gj)//factor[1]
        for mj in range(mj_num):
            jc_num = int((100 - factor[0]*gj - factor[1]*mj)//factor[2])
            if (gj+mj+jc_num) == 100:
                L.append([gj,mj,jc_num])

    for x in L:
        for xx in x:
            print(xx,end=' ')
        print()

    if L !=[]:
        return 0
    else:
        return -1

while True:
    try:
         a = int(input())
         GetResult()

    except:
        break

方法2：python 函数法

while True:
    try:
        n   = int(input())
        # 鸡公最多买20只
        for x in range(21):
            y = (100-7*x)/4 # 鸡母的数量
            z = 100 - x - y # 鸡雏的数量
            if y == int(y) and y >= 0 and z >= 0:
                print(x, int(y), int(z))
    except :
        break