qian = ["["]
hou = ["]"]
adic = {}
for i in range(len(qian)):
adic[qian[i]] = hou[i]
alist = str(input())
n = int(input())
z = 0
for i in alist:
if i in qian:
z += 1
if 1 <= n <= z:
result = []
for i in range(len(alist)):
if alist[i] in qian:
result.append(alist[i])
if len(result) == n:
q = i
break
mubiao = adic[alist[q]]
count1 = 0
for j in range(len(alist)):
if alist[::-1][j] == mubiao:
count1 += 1
if count1 == n:
m = j
break
x = eval(alist[q:len(alist)-m])
y = 0
for i in x:
if isinstance(i,int):
y += 1
print(y)
else:
print(0)
uj5u.com熱心網友回復：
試題如下：輸入一個嵌套串列，再輸入層數，求該層的數字元素個數。
輸入格式:
第一行輸入串列 第二行輸入層數
輸出格式:
在一行中輸出元素個數
輸入樣例:
在這里給出一組輸入。例如：
[1,2,[3,4,[5,6],7],8]
3
輸出樣例:
在這里給出相應的輸出。例如：
2
uj5u.com熱心網友回復：
qian = ["["]
hou = ["]"]
adic = {}
for i in range(len(qian)):
adic[qian[i]] = hou[i]
alist = str(input('陣列'))
n = int(input('層數？'))
z = 0
for i in alist:
if i in qian:
z += 1
if 1 <= n <= z:
result = []
for i in range(len(alist)):
if alist[i] in qian:
result.append(alist[i])
if len(result) == n:
q = i
break
mubiao = adic[alist[q]]
count1 = 0
for j in range(len(alist)):
if alist[::-1][j] == mubiao:
count1 += 1
if count1 == n:
m = j
break
x = eval(alist[q:len(alist)-m])
y = 0
for i in x:
if isinstance(i,int):
y += 1
print(y)
else:
print(0)
沒毛病呀
uj5u.com熱心網友回復：
貌似沒問題啊轉載請註明出處，本文鏈接：https://www.uj5u.com/qita/60817.html
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