天梯赛

前言

网址链接：https://pintia.cn/problem-sets/994805260223102976/exam/problems/type/7?problemSetProblemId=994805324509200384&page=0

1-20

1001 害死人不偿命的(3n+1)猜想

n = int(input())
res = 0
while n!=1:
    if n%2==0:
        n=n/2
    else:
        n=(3*n+1)/2
    res+=1
print(res)

1002 写出这个数

n = input()
number=["ling","yi","er","san","si","wu","liu","qi","ba","jiu"]
num =0
for i in n:
    num = num+int(i)
num = str(num)
res =[]
for index in num:
    res.append(number[int(index)])
print(" ".join(res))

1003 我要通过！

1.条件1就没什么说的，注意不要有非法字符，可以用一个计数器专门记录非法字符，一旦有就是NO。

2.在条件1的基础上，条件2说的是P之前的空字符或者A的数量要等于T之后的空字符或者A字母数量，由于这里P和T只出现一次，因此if判断找到这两个P、T字母之后计算前后数量相等就可以了。

3.条件三说的是P左边一个，PT中间两个，T之后两个。因此条件2三综合就是P左边的A或者空字符数量乘PT中间的数量等于T右边的数量。

n = int(input())

def Solve(input_str):
    num_of_P = input_str.count('P')
    local_of_P = input_str.find('P')
    num_of_T = input_str.count('T')
    local_of_T = input_str.find('T')
    num_of_A = input_str.count('A')
    if num_of_P != 1 or num_of_T !=1 or len(input_string) != num_of_P+num_of_A+num_of_T or local_of_T-local_of_P <=1:
        print('NO')
    elif local_of_P*(local_of_T-local_of_P-1)==len(input_string)-local_of_T-1:
        print('YES')
    else:
        print('NO')

for i in range(n):
    input_string = str(input())
    Solve(input_string)

1004 成绩排名

n = int(input())
students = []

for _ in range(n):
    name, student_id,score= input().split()
    score = int(score)
    students.append((name, student_id, score))
students.sort(key=lambda x: x[2], reverse=True)

print(students[0][0], students[0][1])
print(students[-1][0], students[-1][1])

1005 继续(3n+1)猜想

n = int(input())
numbers = list(map(int, input().split()))
res = []
for num in numbers:
    while num!=1:
        if num%2==0:
            num=num/2
        else:
            num=(3*num+1)/2
        if num not in res:
            res.append(int(num))
key=[]
for num in numbers:
    if num not in res:
       key.append(num)
key.sort(reverse=True)
print(" ".join(map(str, key)))