首先题目大意是:

有两种杯子,一个是5L的容量,一个是3L的容量,如何得到4L的水.(水是无限的,可以适当的文字描述)

答案:

很显而易见是有两种情况

  1. 3+3-5+3
  2. 5+5-3
    显然第二种的方法最好

我试想能不能用代码来实现

目前还有些缺陷....有好的方法可以留言... XD

2020.06.23留，此代码不是最佳选择...

例:


def run(a,b,target):
    result_0 , times_0 = add(a,b,target)
    result_1 , times_1 = sub(a,b,target)
    result = ''
    if times_0 > times_1:
        result = 'RunTimes: {0}\nResult: {1}'.format(times_1,result_1)
    elif times_0 < times_1:
        result = 'RunTimes: {0}\nResult: {1}'.format(times_0,result_0)
    else:
        result = 'RunTimes: {0}\nResult: {1}'.format(times_0,result_0)
    return result

def add(a,b,target):
    box = {}
    box['max_box'] = int(max(int(a),int(b)))
    box['min_box'] = int(min(int(a),int(b)))
    target = int(target)
    if target > box['max_box']:
        return 'False',0
    elif target == box['min_box'] or target == box['max_box']:
        print('run times: 1')
        return target,1
    add = box['min_box']
    box['result'] = count = 0
    result = '{0} = '.format(target)
    while(box['result'] != target):
        box['result'] += int(add)
        result  += ' +{} '.format(add)
        count += 1
        if box['result'] >= box['max_box'] :
            box['result'] -= box['max_box']
            result += ' -{} '.format(box['max_box'])
            count += 2 #如果算上倒水的 则改为 2
    return result , count

def sub(a,b,target):
    box = {}
    box['max_box'] = int(max(int(a),int(b)))
    box['min_box'] = int(min(int(a),int(b)))
    target = int(target)
    if target > box['max_box']:
        return 'False',0
    elif target == box['min_box'] or target == box['max_box']:
        print('run times: 1')
        return target,1
    sub = box['min_box']
    box['result'] = box['max_box']
    #box['tmp'] 剩余容量的变量 
    box['tmp']  = count = 0
    result = '{0} ='.format(target)
    while(box['result'] != target):
        result += ' +{0} -{1}'.format(box['result'],sub)
        box['result'] -= sub
        count += 1
        if box['result'] > 0 :
            result += ' +{0} -{1} + {2}'.format(box['max_box'],box['min_box'],box['result'])
            box['result'] = box['max_box'] - box['min_box'] + box['result']
            count += 2 #如果算上倒水的 则改为 2
    return result , count

a,b,target = input('Input a b target: ').split()
print(run(a,b,target))