Coding Everyday
2019/10/12
洛谷【P1425】
#wrong
time = input().split(" ")
hours = int(time[2])-int(time[0])
minutes = 60-int(time[1]) + int(time[3])
print(str(hours-1)+" "+str(minutes))
- 统一单位之后一切都好办
#right
time = input().split(" ")
first = 60*int(time[0]) + int(time[1])
second = 60*int(time[2]) + int(time[3])
minus = second - first
#
print(str(minus//60),str(minus%60))
#
2019/12/13
Lintcode【Trailing Zeros】
- 计算0的个数可以换成计算2和5的个数，进而转化成算5的个数
def trailingZeros(self,n):
# write your code here, try to do it without arithmetic operators.
num = 0
while n>0:
num = num + n//5
n = n//5
return num
2019/12/14
LintCode【Digit Counts】
- 将数字转化成字符串进行处理
def digitCounts(self, k, n):
# write your code here
#
str_k = str(k)
#
count = 0
for i in range(n+1):
for j in str(i):
if j == str_k:
count = count+1
return count
2019/12/15
PAT【1001 害死人不偿命的(3n+1)猜想】
n = int(input())
times = 0
#
if n == 1:
print(0)
#
else:
while n != 1:
if n%2 == 0:
n = n/2
times +=1
else:
n = (3*n+1)/2
times +=1
print(times)
PAT【1002 写出这个数】
- 数字转化成字符串处理
num = input()
sum = 0
d = {"1":"yi","2":"er","3":"san","4":"si","5":"wu","6":"liu","7":"qi","8":"ba","9":"jiu","0":"ling"}
for i in num:
sum += int(i)
str_sum = str(sum)
iteration_times = len(str_sum)-1
#格式化输出
for j in range(iteration_times):
#int不能迭代。。变量名iteration自然而然以为是迭代了
print(d[str_sum[j]],end=" ")
print(d[str_sum[-1]],end="")
#
#现在看看自己之前写的真的很搞笑
num = input()
sum = 0
d = {1:"yi",2:"er",3:"san",4:"si",5:"wu",6:"liu",7:"qi",8:"ba",9:"jiu",0:"ling"}
for i in num:
number = number + int(i)
if number//100 != 0 :
hundred = number//100
else:
hundred = 0
if (number-hundred*100)//10 != 0 :
decade = (number-hundred*100)//10
else:
decade = 0
if (number-hundred*100-decade*10) != 0 :
digit = number-hundred*100-decade*10
else:
digit = 0
if hundred:
print(d[hundred],end=" ")
if decade:
print(d[decade],end=" ")
if digit:
print(d[digit])
else:
print(d[0])
PAT【1003 我要通过！】
正则表达式
#之前写的，总是非零返回。。
import re
str_to_be_match = input().split("\n")
for i in range(int(str_to_be_match[0])):
if re.match(r'(A*)P(A+)T(A*)',str_to_be_match[i+1]):
print("YES")
else:
print("NO")
PAT【1006 换个格式输出整数】
#之前写的
num = int(input())
if num // 100 != 0:
hunderd = num // 100
else:
hunderd=0
if (num-100*hunderd)//10 != 0:
decade = (num-100*hunderd)//10
else:
decade = 0
if (num-hunderd*100-decade*10) != 0:
digit = num-hunderd*100-decade*10
else:
digit = 0
for i in range(hunderd):
print("B",end="")
for i in range(decade):
print("S",end="")
for i in range(digit):
print(i+1,end="")
- 判断数字的位数同样可以通过将数字转化成字符串处理
num = input()
if len(num)==3:
print(int(num[-3])*"B",end="")
if len(num)>=2:
print(int(num[-2])*"S",end="")
if num[-1]:
for i in range(int(num[-1])):
print(i+1,end="")
LeetCode【最长有效括号】
#只筛选出来了包含成对（）的字符串；下次继续改
class Solution:
def longestValidParentheses(self, s: str) -> int:
ls = []
num = 0
for i in s:
if i == "(":
ls.append(i)
else:
if ls:
num+=1
ls.pop()
return num*2
# 只筛选出来了连续出现的（）的字符串；嵌套的不能筛选出来，很失败，下次继续改
def longestValidParentheses(self, str_to_bematched: str) -> int:
ls = []
num = 0
left_of_symbol_exist = False
for i in str_to_bematched:
if i == "(" and left_of_symbol_exist == False:
ls.append(i)
left_of_symbol_exist = True
if i == ")" and left_of_symbol_exist == True:
if ls:
num+=1
ls.pop()
left_of_symbol_exist = False
return num*2
考试周，没时间搞CS
2019/12/27
#用python仿造栈数据结构
class Stack:
def __init__(self):
self.items() = []
def push(self,item):
self.items.append()
def isEmpty(self):
return self.items == []
def pop(self,item):
return self.items.pop()
def size(self):
return len(self.items)