如何在同一行中给出元素的数量和列表?下面是一个例子。
234 567 321 345 123 110 767 111在这里,我想问用户他/她想要多少个元素,也就是N,在指定之后,我想让用户在他/她输入N的同一行中输入元素。
T = int(input())
for x in range(T):
N = int(input())
nums = list(map(int, input().split()))
在这段代码中,我可以在不同的行中给出N和nums,但我希望它们在同一行中,然后评估解决方案。
以下是完整的代码。
def largest(num):
num_str = str(num)
i = 9
while i >= 0:
if str(i) in num_str:
return i
i -= 1
def smallest(num):
num_str = str(num)
i = 0
while i <= 9:
if str(i) in num_str:
return i
i += 1
def pairs(num):
if num == 2:
return 1
if num == 3:
return 2
return 0
T = int(input())
for x in range(T):
N = int(input())
nums = list(map(int, input().split()))
assert(len(nums) == N)
bitscore = [""]*N
for i in range(len(bitscore)):
curr_score = str(11 * largest(nums[i]) + 7 * smallest(nums[i]))
if len(curr_score) > 2:
curr_score = curr_score[-2:]
bitscore[i] = curr_score
odd_pos_freq = [0] * 10
even_pos_freq = [0] * 10
for i in range(len(bitscore)):
index = int(bitscore[i][0])
if (i + 1) % 2 == 0:
even_pos_freq[index] += 1
else:
odd_pos_freq[index] += 1
count_pairs = [0] * 10
for i in range(10):
if even_pos_freq[i] <= 1 and odd_pos_freq[i] <= 1:
continue
count_pairs[i] += pairs(even_pos_freq[i]) + pairs(odd_pos_freq[i])
count_pairs[i] = min(2, count_pairs[i])
print(sum(count_pairs))
我知道这个问题可能听起来很愚蠢,但很多时候我都被这种小事卡住了,即使有正确的逻辑,我也无法提交正确的答案。