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题解 | #字符串分隔#

import sys for line in sys.stdin: line = line.split('\n')[0] # 求余，不足补0 if len(line)%8!=0: line = line + (8-(len(line)%8))*'0' ...

2023-04-09

0 244

py

python Letters = 0 Digits = 0 Others = 0 x = input() for i in x: if i.isdigit(): Digits += 1 elif i.isalpha(): Letters += 1 ...

2020-09-24

2 392

py

python def diognose(): # 要判断的数是 x = int(input()) flag = 0 for i in range(1, x + 1): if x % i == 0: flag += 1 ...

2020-09-24

0 398

爆破

x =int(input()) y=[] for j in range(0, x+1): for k in range(0, x+1): for l in range(0, x+1): if j+2*k+5*l==x: ...

2020-09-16

0 443

Python

#输出1~n中能被3整除，且至少有一位数字是5的所有整数. #思路：题中两个条件还是要用列表先拆分数位，数位和要对3整除。用上一个题的模板。 x = int(input()) #首先生成1-x的整数 for i in range(1,x+1): y = len(str(i)) if...

2020-09-16

2 450

This

python # 思路一：整数转列表列表切片reverse，原列表等于倒序列表则为回文复杂度o（n^2） #输入整数 x = int(input()) #首先生成1-x的整数 for i in range(1,x+1): y = len(str(i)) if y == 1: ...

2020-09-16

0 406

This

py 输入 x = float(input())PI = 3.14159#计算out1 = 2xPIout2 = PIxx#输出print("%0.f"%(6,out1))print("%0.f"%(6,out2))

2020-09-10

0 463

This

输入三个数 x1 = list(input().split())#三数据处理length = float(x1[0])weight = float(x1[1])height = float(x1[2])#计算公式x = float(height(length + weight))/2print("%...

2020-09-10

0 402

This

py#输入x1 = list(input().split())x2 = list(input().split()) 计算公式 a = int(x1[0])b = int(x1[1])c = int(x2[0])d = int(x2[1])x = (c-a)*60+d-b #输出 print(x)

2020-09-10

0 475