我正在尝试解决这个 CodingBat 问题：
我们想做一包目标公斤的巧克力。我们有小条（每条 1 公斤）和大条（每条 5 公斤）。返回要使用的小柱的数量，假设我们总是在小柱之前使用大柱。如果无法完成，则返回 -1。
我理解问题的逻辑，但每当我尝试运行它时，都会出现超时异常。那么谁能告诉我我做错了什么？
def make_chocolate(small, big, goal):
total = 0
if goal < 5:
big = 0
for i in xrange(big):
total += 5
if total == goal:
return 0
elif total+5>goal:
break
for k in xrange(small):
total +=1
if total == goal:
return (k+1)
return -1
失败是由执行此测试所需的时间引起的：
makeChocolate(1000, 1000000, 5000006)
即使其他测试可以通过，一旦一个测试超时，报告会将所有测试显示为超时。要看到这是真的，请更改xrange(big)为xrange(big if big < 101 else 0)，您将看到除上述测试之外的所有测试通过。
出于性能原因，基于 Web 的评估器需要像这样的超时。一定是 Python 超时允许的循环少于 Java 超时。
这是通过的非循环解决方案：
def make_chocolate(small, big, goal):
big *= 5
if big + small < goal or small < goal%5:
return -1
small = goal - big
return small%5 if small < 0 else small
Java 需要一个稍微不同的解决方案，因为它如何处理负数的模数。
public int makeChocolate(int small, int big, int goal) {
big *= 5;
if (big + small < goal || small < goal%5)
return -1;
small = goal - big;
return small < 0 ? (big+small)%5 : small;
}
你可以简单地这样做：
def make_chocolate(small, big, goal):
noOfBigs = big if(5 * big <= goal) else goal / 5
return goal - (noOfBigs * 5) if small >= (goal - (noOfBigs * 5)) else -1
如果您首先将 Big value 减少到您实际使用的数字，这似乎是更清晰的逻辑：
def make_chocolate(small, big, goal):
while big * 5 > goal:
big -= 1
if (goal - (big * 5)) <= small:
return goal - (big * 5)
else:
return -1
This is clunky, but it works:
def make_chocolate(small, big, goal):
bigbars=goal//5
if bigbars<=big:
smallgoal=goal-(bigbars*5)
if smallgoal>=0 and smallgoal<=small:
return smallgoal
if smallgoal>small:
return -1
if bigbars>big:
smallgoal=goal-(big*5)
if smallgoal<=small:
return smallgoal
if smallgoal>small:
return -1
def make_chocolate(small, big, goal):
big = big*5
if (goal >= big) and (small >= goal - big):
return goal - big
if (goal < big) and (small >= goal % 5):
return goal % 5
return -1
def make_chocolate(small,big,goal):
if goal > small + big * 5:
return -1
else:
if goal % 5 <= small:
if big * 5 <=goal:
return goal - (big * 5)
else:
return goal % 5
else:
return -1
def make_chocolate(small, big, goal):
i = goal // 5
b_big = big*5
x = i*5
y = goal - x
if big == 0:
if small >= goal:
return small - (goal - small)
else:
return -1
elif b_big == goal:
return 0
elif big >= i:
if x + small < goal:
return -1
else:
return (goal - (x))
else:
if b_big + small < goal:
return -1
else:
return(goal - b_big)
def make_chocolate(small, big, goal):
if(goal>=big*5) and(small+big*5>=goal) and (goal%5 <=small):
rem_cho=abs(goal-big*5)
return(rem_cho)
elif(goal%5>small):
return(-1)
elif(goal<big*5):
return(goal%5)
else:
return(-1)
以下代码适用于所有测试数据。对不起，由于它是自我解释的，我没有提供任何解释。
def make_chocolate(small, big, goal):
s = small*1
b = big*5
mod = goal%5
if b < goal:
if s+b == goal:
return s
elif s+b > goal:
return goal-b
elif s+b < goal:
return -1
elif b > goal:
if s < mod:
return -1
return mod
elif b == goal:
return 0
def make_chocolate(small, big, goal):
can_do = small + 5*big
if can_do<goal:
return -1
elif can_do==goal:
return small
else:
b_max = goal/5
b_max = b_max if b_max<=big else big
s_req = goal - (b_max)*5
if s_req <= small:
return s_req
else:
return -1
def make_chocolate(small, big, goal):
s_cnt = 0;
t = goal;
for i in range(**550**):
if(goal >= 5 and big > 0):
goal = goal-5;
big = big-1;
elif(goal > 0 and small > 0):
goal = goal-1;
small = small-1;
s_cnt = s_cnt+1;
if (goal == 0):
return s_cnt;
else:
return -1;
// 在大写字母范围（550）中，您可以看到我已经对其进行了硬编码，因为对于较大的值，它将使循环运行时间过长，因此您会收到超时错误。
def make_chocolate(small, big, goal):
if goal - 5 * big == 0:
return 0
elif (goal - 5 * big) < 0:
return -1 if (goal % 5) > small else goal % 5
elif (goal - 5 * big) > small:
return -1
else:
return (goal - 5 * big)
def make_chocolate(small, big, goal):
while big*5>goal:
if goal%5==0:
return 0
else:
if goal%5<=small:
return goal%5
else:
return -1
else:
if goal-(big*5)<=small:
return goal-(big*5)
else:
return -1
def permutations(amount, size, threshold):
if amount == 1:
return size
combinations = []
for i in range(1, amount + 1):
combinations.append(size * i)
if combinations[-1] == threshold:
return combinations[-1]
elif combinations[-1] > threshold:
if size == 1:
return combinations[-1] - threshold
elif size == 5:
return combinations[-2]
else:
return combinations
def make_chocolate(small_amount, big_amount, goal):
"Adjusts values for small and big chocolate bars until we find out whether there is a solution"
if big_amount == 0 and small_amount == goal:
return goal
elif big_amount == 0 and small_amount != goal:
return -1
else:
big_size = 5
small_size = 1
big_sums_list = permutations(big_amount, big_size, goal)
big_sum = big_sums_list
if type(big_sum) == list: #If none of the sums reached our goal
new_threshold_value = goal - big_sums_list[-1] # We adjust the new threshold value for the permutations function
small_sums_list = permutations(small_amount, small_size, goal) #Get all possible sums with our given small values
small_sum = small_sums_list # If the sum of our small values has indeed equaled or exceeded our goal
if type(small_sums_list) == list: # If the sum of our small values has not equaled nor exceeded our goal
small_sums_list = permutations(small_amount, small_size, new_threshold_value)
small_sum = small_sums_list
if type(small_sums_list) == list: # If the sum of small bars has not exceeded the threshold value
return -1
elif small_sum == new_threshold_value:
return small_sum # Whichever value we choose will be correct
elif small_sum > new_threshold_value:
return new_threshold_value # It is the number of small bars we have left before our goal that matters
elif small_sum == goal:
return small_sum # If the sum has equaled our goal, it is indiferent whether we choose one or the other variable
elif small_sum > goal: # If the sum has exceeded our goal
return goal #Since the size is one, we can return goal as the result
elif big_sum >= goal:
return make_chocolate(small_amount, big_amount - 1, goal) # We adjust the amounts by - 1, to check for different combinations
elif big_sum == 5:
new_threshold_value = goal - big_sum # We adjust the new threshold value for the permutations function
small_sums_list = permutations(small_amount, small_size, new_threshold_value) #Get all possible sums with our given small values
small_sum = small_sums_list # If the sum of our small values has indeed equaled or exceeded our goal
if type(small_sum) == list: # If the sum of our small values has not equaled nor exceeded our goal
return -1
elif small_sum == new_threshold_value:
return small_sum # If the sum has equaled our goal, it is indiferent whether we choose one or the other variable
elif small_sum > new_threshold_value: # If the sum has exceeded our goal
return goal #Since the size is one, we can return goal as the result
print(make_chocolate(4, 1, 10))