def sum100(n):
if n == 1:
return 1
else:
return n + sum100(n-1)
print(sum100(100))
if n == 1:
return 1
else:
return n + sum100(n-1)
print(sum100(100))
2023-06-26
def sub_sum(L):
a = 0
sum_1 = 0
sum_2 = 0
for a in L:
if a % 2 == 0:
sum_1 = sum_1 + a
else:
sum_2 = sum_2 + a
return sum_2,sum_1
L = [1,2,3,4,5,6,7,8]
sum_1,sum_2 = sub_sum(L)
print('单数项的和:{}'.format(sum_2))
print('偶数项的和:{}'.format(sum_1))
a = 0
sum_1 = 0
sum_2 = 0
for a in L:
if a % 2 == 0:
sum_1 = sum_1 + a
else:
sum_2 = sum_2 + a
return sum_2,sum_1
L = [1,2,3,4,5,6,7,8]
sum_1,sum_2 = sub_sum(L)
print('单数项的和:{}'.format(sum_2))
print('偶数项的和:{}'.format(sum_1))
2023-06-26
# Enter a code
L = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
S = set([1, 3, 5, 7, 9, 11])
for a in L:
if a in S:
S.remove(a)
else:
S.add(a)
print(S)
L = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
S = set([1, 3, 5, 7, 9, 11])
for a in L:
if a in S:
S.remove(a)
else:
S.add(a)
print(S)
2023-06-25
# Enter a code
d = {
'Alice': 45,
'Bob': 60,
'Candy': 75,
'David': 86,
'Ellena': 49
}
score = []
for name in d:
print(name)
score.append(d[name])
for key in list(d.keys()):
if key in d:
d.pop(key)
print(score)
print(d)
d = {
'Alice': 45,
'Bob': 60,
'Candy': 75,
'David': 86,
'Ellena': 49
}
score = []
for name in d:
print(name)
score.append(d[name])
for key in list(d.keys()):
if key in d:
d.pop(key)
print(score)
print(d)
2023-06-24
d = {
'Alice': 45,
'Bob': 60,
'Candy': 75,
'David': 86,
'Ellena': 49
}
if 'Alice' in d:
score = d['Alice']
print(score)
d['Alice'] = 60
print(d)
'Alice': 45,
'Bob': 60,
'Candy': 75,
'David': 86,
'Ellena': 49
}
if 'Alice' in d:
score = d['Alice']
print(score)
d['Alice'] = 60
print(d)
2023-06-24
ju = [[1,2,3],[5,3,2],[7,3,2]]
print(ju)
a = 0
for a in range(len(ju)):
print(ju[a][0]*ju[a][1]*ju[a][2])
a += 1
print(ju)
a = 0
for a in range(len(ju)):
print(ju[a][0]*ju[a][1]*ju[a][2])
a += 1
2023-06-23
L = ['Alice', 'Chinese', 92, 'Math', 75, 'English', 99]
print(L)
print(L)
2023-06-16
set提供isdisjoint()方法,可以快速判断两个集合是否有重合,如果有重合,返回False。##竟然记反了
2023-06-12