您好，我已經用我的代碼更新了此內容：輸入->a=[16,21, 56, 40]def find_primes(a): num_factors=[] for num in a: # print('1',num) list_of_factors=[] i=2 while num>1: #print('2',num) if num%i==0: list_of_factors.append(i) # print('3',list_of_factors) num=num/i # print('4',num) i=i-1 i+=1 num_factors.append(list_of_factors) d = {} a_list = [] for i in num_factors: for j in i: d[j] = d.get(j, 0) + 1 dictionary_copy = d.copy() a_list.append(dictionary_copy) print(a_list) return num_factors find_primes(a)這是我得到的輸出：[{2: 10, 3: 1, 7: 2, 5: 1}][[2, 2, 2, 2], [3, 7], [2, 2, 2, 7], [2, 2, 2, 5]]這是我的問題：理解因為它是一本字典，所以鍵的值會累積。我想計算每個列表中唯一整數的數量。例如。[{2:4},{3:1,7:1},{2:3,7:1},{2:3,5:1}] 而不是上面代碼的輸出中給出的內容。之后，我想獲得每個整數的最大出現次數來計算 LCM。2^4 * 3 * 7 * 5請建議我們如何改進代碼。感謝您的幫助。
1 回答
qq_花開花謝_0
TA貢獻1835條經驗 獲得超7個贊
這可能是其中一種方法。
numbers=list(map(int,input().split()))
numbers.sort()
maximum=numbers[0] #gcd of the input numbers cannot be greater than minimum number in the list.Therefore we are retrieving the mininum number of the list.
gcd=1 #Initial value of gcd.
for i in range(1,(maximum+1)):
flag=0
for j in range(len(numbers)):
num=numbers[j]
if num % i ==0: #check if the every number in the input is divisible by the value of i
flag=1 #if divisible set flag=1 and then check the other numbers of the input.
else:
flag=0 #if any of the number of the input is not divisible then flag =0 and go to next iteration of i.
break
if flag==1:
gcd=i #if flag=1 that means every number of the input is divisible by the value of j.Update the value of gcd.
print(gcd)
可以通過以下方式完成：
for i in num_factors:
d={}
for j in i:
try:
d[j]=d[j]+1
except KeyError:
d[j]=1
dictionary_copy = d.copy()
a_list.append(dictionary_copy)
print(a_list)
return num_factors
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