import randomn = random.randint(1,100)if (n % 2 != 0): print("Weird")elif (n % 2 == 0 and n in range(2,5)): print("Not Weird")elif (n % 2 == 0 and n in range(6,20)): print("Weird")elif (n % 2 == 0 and n > 20): print("Not Weird") 這段代碼有問題嗎？
2 回答
Cats萌萌
TA貢獻1805條經驗 獲得超9個贊
您的范圍檢查并不嚴格，例如：包括andrange(2, 5)之間的整數，但無論如何有一種更簡單的方法來編寫它；嘗試這個：24
import random
n = random.randint(1,100)
if n % 2 != 0:
print("Weird")
elif 2 <= n <= 5: # from this point on, the number can only be even
print("Not Weird")
elif 6 <= n <= 20:
print("Weird")
elif n > 20:
print("Not Weird")
或者更好的是，它更簡單并且做同樣的事情：
if n % 2 != 0 or 6 <= n <= 20:
print("Weird")
else:
print("Not Weird")
富國滬深
TA貢獻1790條經驗 獲得超9個贊
隨機導入
n = random.randint(1,100)
print(n)
if (n % 2 != 0):
print("Weird")
elif (n % 2 == 0 and n in range(2,5)):
print("Not Weird")
elif (n % 2 == 0 and n in range(5,20)):
print("Weird")
elif (n % 2 == 0 and n >= 20):
print("Not Weird")
您缺少一些范圍，我剛剛更新了代碼嘗試一下。
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