我有一個(gè)while循環(huán)，該循環(huán)返回爬山所需的“奔波”次數(shù)。小山的大小是“斜坡高度”，爬升的高度是“ rush_height_gain”減去“ back_sliding”。以下代碼適用于：ans = num_rushes(15, 10, 5)print(ans)打印1和ans = num_rushes(100, 15,7)print(ans)打印2和ans = num_rushes(10, 10, 9)print(ans)打印12但是返回錯(cuò)誤的答案ans = num_rushes(100, 10, 0)print(ans)應(yīng)該打印10，但是打印9我不確定為什么會這樣，任何幫助將不勝感激def num_rushes(slope_height, rush_height_gain, back_sliding): current_height = 0 rushes = 0 while current_height < slope_height: if rush_height_gain == slope_height: rushes+=1 return rushes elif current_height < slope_height: if current_height == slope_height: return rushes else: a = rush_height_gain - back_sliding current_height += a if current_height == slope_height: return rushes elif current_height > slope_height: return rushes else: rushes+=1 return (rushes)
4 回答
慕斯709654
TA貢獻(xiàn)1840條經(jīng)驗(yàn) 獲得超5個(gè)贊
如果我正確地理解了這個(gè)問題，那么我認(rèn)為您正在尋找的是：
def num_rushes(slope_height, rush_height_gain, back_sliding):
if rush_height_gain < slope_height and rush_height_gain - back_sliding < 1:
raise Exception("this is not going to work very well")
current_height = rushes = 0
while current_height < slope_height:
rushes += 1
current_height += rush_height_gain
if current_height >= slope_height:
break
current_height -= back_sliding
return rushes
每次上坡“奔波”后，您都要檢查是否已經(jīng)到達(dá)山頂。如果是這樣，那么您就完成了；如果沒有，那么請向下滑動一下再走一次！正如@perreal在他對原始帖子的評論中的鏈接中所指出的那樣，如果您滑倒的次數(shù)多于滑倒的次數(shù)，并且第一次沒有完全站起來，那么您將會遇到問題。在這些情況下，您可能想拋出一個(gè)異常。
犯罪嫌疑人X
TA貢獻(xiàn)2080條經(jīng)驗(yàn) 獲得超4個(gè)贊
我相信問題是這樣的：
if current_height == slope_height:
return rushes
如果back_sliding是0，那么在第十次迭代，current_height從去90到100。然后，支票返回true，并9在遞增之前返回。
呼如林
TA貢獻(xiàn)1798條經(jīng)驗(yàn) 獲得超3個(gè)贊
def num_rushes(slope_height, rush_height_gain, back_sliding):
current_height = 0
rushes = 0
while current_height < slope_height:
current_height += rush_height_gain
rushes += 1
if current_height < slope_height:
current_height -= back_sliding
return rushes
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