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#Familiarizing text editor,IDE,code analysis tools.
print("Enter your address")
n=input("Enter your name : ")
hname=input("Enter your house name : ")
p=input("Enter your place : ")
pin=int(input("Enter your pincode: "))
d=input("Enter your district : ")
print("Address details :")
print("Name :",n)
print("House... |
#Program to find the factorial of a number
a=int(input("Enter the number:"))
fact=1
for x in range(1,a+1):
fact=fact*x
print("The factorial",fact)
|
#Generate a list of four digit numbers in a given range with all their digits even and the number is a perfect square.
a=int(input("Enter the lower digit:"))
b=int(input("Enter the upper digit:"))
import random
print(random.randint(a,b))
c=[]
c=[x for x in range(a,b)if(int(x**0.5)**2==x and int(x%2==0))]
print(c... |
#List of color names entered by user and display first and last color
c=[]
x=int(input("Enter limit: "))
for i in range(1,x+1):
a=input()
c.append(a)
print(c)
print(c[0],c[-1])
|
#Program to check whether a number is positive or negative
a=int(input("Enter the number:"))
if a>0:
print("The given number is positive")
elif a==0:
print("The given number is zero")
elif a<0:
print("The given number is negative")
|
#Program to check whether the given number is palindrome or not
a=int(input("Enter the number:"))
temp=a
rev=0
while a > 0:
d=a%10
rev=rev*10+d
a//=10
if temp==rev:
print("The given number is Palindrome")
else:
print("The given number is not palindrome")
|
class BinarySearch(list):
def __init__(self,a,b):
self.a = a # a is length of ist to be created
self.b = b # b is the step between consequtive numbers
self.li = [i for i in range(b,a+1)]
self.length = len(self.li)
def search(self,item):
self.count =0
self.ind... |
def remove_parentheses(s):
level = {s}
while True:
res = []
for cur in level:
if is_valid(cur):
res.append(cur)
if res:
return res
new_level = set()
for cur in level:
for i in range(len(cur)):
new_level... |
def max_swap(num):
digits = list(str(num))
res = 0
for i in range(len(digits)-1):
for j in range(i+1, len(digits)):
c_digits = digits[:]
c_digits[i], c_digits[j] = c_digits[j], c_digits[i]
if int("".join(c_digits)) > res:
res = int("".join(c_digits... |
from typing import List
from collections import defaultdict
from queue import Queue
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Solution:
def verticalTraversal(self, root: TreeNode) -> List[List[int]]:
... |
from typing import List
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Solution:
def rob(self, root):
cache = {}
return self._rob(root, cache)
def _rob(self, root: TreeNode, cache) -> int... |
# dummy class
class BinaryMatrix(object):
def get(self, row: int, col: int) -> int:
return 1
def dimensions(self) -> list:
return []
class Solution:
def leftMostColumnWithOne(self, binaryMatrix: 'BinaryMatrix') -> int:
n_row, n_col = binaryMatrix.dimensions()
j = n_col - ... |
from typing import List, Set
class Solution:
def count_components(self, n, edges: List[List[int]]) -> int:
if n == 0:
return 0
graph = self._create_adjacent_list(n, edges)
visited = set()
res = 0
for v in range(n):
if not v in visited:
... |
def max_diff(root):
res = [0]
dfs(root, 0, 100000, res)
return res[0]
def dfs(node, max_val, min_val, res):
"""
Top-down approach. Keep track of the max and min encountered when going down.
When reaching a leaf, the max diff is max_val - min_val. Since dfs covers all the
path, the max diff... |
from collections import Counter
import heapq
class Word:
def __init__(self, w, c):
self.w = w
self.c = c
def __lt__(self, other):
if self.c < other.c:
return True
elif self.c == other.c:
return self.w > other.w
else:
return False
... |
from queue import Queue
def is_bipartite(graph):
colors = dict() # store nodes and its color
q = Queue()
for i in range(len(graph)):
if i not in colors:
q.put(i)
colors[i] = 0
while not q.empty():
cur = q.get()
for v in graph[c... |
from collections import deque
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
"""
Using the BST property, we can the value range to determine if the current
node is done. Comparing with serialize and deserialize a BT, we don't need to
add indicator ... |
import heapq
from typing import List
class Solution:
def k_smallest_pairs_brutal_force(self, nums1: List[int], nums2: List[int], k: int) -> List[List[int]]:
"""
Time complexity O(m*n*log(k))
Space complexity O(k)
"""
if not nums1 or not nums2:
return []
... |
from typing import List
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Solution:
"""
Flatten the tree using in-order traversal and check if the array is sorted.
"""
def isValidBST(self, root: Tre... |
from typing import List
class Solution:
def length_of_lis(self, nums: List[int]) -> int:
"""
maintain a array dp whose i-th element is the length of the longest increasing substring
in nums[:i+1] and must contain nums[i], the dynamic programming relation is:
dp[i] = max(dp[j]) + 1... |
# Definition for a binary tree node.
from queue import Queue
class TreeNode(object):
def __init__(self, x):
self.val = x
self.left = None
self.right = None
class CodecPreOrder:
def serialize(self, root):
"""
Use pre-order traversal to serialize the tree
"""
... |
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
class Solution:
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
cur1 = l1
cur2 = l2
dummy = ListNode(0)
cur = dummy
carry = 0
while cur1 or ... |
def product(nums):
front = [1]
for num in nums[:-1]:
front.append(front[-1] * num)
end = [1]
for num in nums[::-1][:-1]:
end.append(end[-1] * num)
end = end[::-1]
res = []
for i in range(len(nums)):
res.append(front[i] * end[i])
return res
def product_space_... |
def is_sorted(words, order):
n_order = 'abcdefghijklmnopqrstuvwxyz'
map_dict = {}
for c1, c2 in zip(order, n_order):
map_dict[c1] = c2
n_words = []
for w in words:
temp = []
for c in w:
temp.append(map_dict[c])
n_words.append(''.join(temp))
for i in... |
from typing import List
class Solution:
def max_sub_array(self, nums: List[int]) -> int:
"""
maintain an array dp and dp[i] is the maximum subarray that ends with nums[i].
Thus:
dp[i+1] = nums[i+1] if dp[i] <= 0
= nums[i+1] + dp[i] if dp[i] > 0
"""
... |
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next # class ListNode:
def partition(head, x):
left_head = ListNode(None)
right_head = ListNode(None)
left = left_head
right = right_head
cur = head
while cur:
if cur.val < x:
... |
class Solution:
def insert(self, intervals, new_interval):
res = []
new_low, new_high = new_interval
for i in range(len(intervals)):
if intervals[i][1] < new_low:
res.append(intervals[i])
elif intervals[i][0] > new_high:
res.append([ne... |
def is_equal(nums):
if len(nums) < 7:
return False
n_sum = [nums[0]]
for i in range(1, len(nums)):
n_sum.append(n_sum[-1] + nums[i])
for j in range(3, len(nums)-3):
# it is possible there are more than one sum value
# 2, 1, -1, 2, 1, remove -1, we have 3, remove 1, we h... |
#!/bin/python
name = "Alli"
print(name)
if len(name) >=5:
print("name is long")
elif len(name) == 4:
print("name is average")
else:
print("name is short")
print("Condition successul")
|
#16진수로 입력된 정수 한 개를 8진수로 바꾸어 출력하는 프로그램을 작성해보자.
#따로 생각나는 방법이 없어서 일단 10진수로 변경후, 다시 8진수로 변경하는 방법을 사용하였다.
num = int(input(), 16)
nums = oct(num)[2:]
print(nums)
'''
다른 사람의 풀이
a=input()
n=int(a, 16)
print('%o' % n)
다음엔 꼭 이 방법으로 풀어봐야지..
'''
|
#정수가 순서대로 입력된다. 0이 아니면 입력된 정수를 출력하고 0이 입력되면 출력을 중단해보자.
#이전에 있던 문제랑 무엇이 다른지요..?
#아무튼 다시 풀었습니다.
a = input().split()
for i in range(len(a)):
if a[i] != '0':
print(a[i])
else:
print(a[i])
break
|
#단어를 하나 입력받는다. 입력받은 단어(영어)의 각 문자를 한줄에 한 문자씩 분리해 출력한다.
#'B''o''y' 이런 식으로 문제에서는 답을 원해서.. 어떤 방법으로 할까 하다가.. 그냥 프린트에 ""를 포함해서 출력해주었다.
#다른 방버이 있었을 지도..
word = list(input())
for i in range(len(word)):
print("'"+word[i]+"'")
|
#두 정수 사이의 합
#https://programmers.co.kr/learn/courses/30/lessons/12912
#두 정수 a, b가 주어졌을 때 a와 b 사이에 속한 모든 정수의 합을 리턴하는 함수, solution을 완성하세요.
#예를 들어 a = 3, b = 5인 경우, 3 + 4 + 5 = 12이므로 12를 리턴합니다.
def adder(a, b):
sum = 0
if a < b:
for i in range (a, b+1):
sum += i
else:
for i in ra... |
#정수 한 개가 입력되었을 때, minus(음)/plus(양) even(짝)/odd(홀)을 출력해보자.
#첫 줄에 minus(음) 나 plus(양) 를 출력하고, 두번째 줄에 odd(홀) 나 even(짝) 을 출력한다.
num = int(input())
if num > 0:
print("plus")
else:
print("minus")
if num%2 == 0:
print("even")
else:
print("odd")
|
#입력된 두 정수 a, b 중 큰 값을 출력하는 프로그램을 작성해보자.
#(단, 조건문을 사용하지 않고 3항 연산자 ? 를 사용한다.)
#삼항 연산자: 항이 3개인 형태일 때
a, b = input().split(" ")
a = int(a)
b = int(b)
c = a if a>b else b
print(c)
#파이썬의 3항 연산자
#(조건을 충족할 때) if (맞는 조건) else (조건을 충족하지 않을 때)
#ex: a if a>b else b
|
# 나의 풀이
def solution(s):
answer = ''
numbers_english = ['zero', 'one', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight', 'nine']
curr = ''
for i in s:
if i.isalpha():
curr += i
if curr in numbers_english:
answer += str(numbers_english.ind... |
#정수 한 개가 입력되었을 때 카운트다운을 출력해보자.
num = int(input())
while num>0:
print(num)
num = num -1
|
#x만큼 간격이 있는 n개의 숫자
#https://programmers.co.kr/learn/courses/30/lessons/12954
#함수 solution은 정수 x와 자연수 n을 입력 받아, x부터 시작해 x씩 증가하는 숫자를 n개 지니는 리스트를 리턴해야 합니다.\
#다음 제한 조건을 보고, 조건을 만족하는 함수, solution을 완성해주세요.
#나의 풀이
def solution(x, n):
answer = []
for i in range(1, n+1):
answer.append(x*i)
return answer... |
# 큰 수 만들기 - 그리디 알고리즘
# https://programmers.co.kr/learn/courses/30/lessons/42883
def solution(number, k):
stack = [number[0]]
for num in number[1:]:
while len(stack) > 0 and stack[-1] < num and k > 0:
k -= 1
stack.pop()
stack.append(num)
return ''.join(stac... |
#정수 두 개(a, b)를 입력받아 합, 차, 곱, 몫, 나머지, 나눈값을 자동으로 계산해보자.
#정수 두 개(a, b)가 공백을 두고 입력된다.
'''
출력 기준
첫 줄에 합
둘째 줄에 차(a-b)
셋째 줄에 곱,
넷째 줄에 a를 b로 나눈 몫,
다섯째 줄에 a를 b로 나눈 나머지,
여섯째 줄에 a를 b로 나눈값(실수, 소수점 셋째 자리에서 반올림해 둘째 자리까지 출력)
을 출력한다.
'''
a, b = input().split(" ")
a = int(a)
b = int(b)
print(a+b)
print(a-b)
print(a*b)
print(a//b)
pr... |
#문자열 내 마음대로 정렬하기
#https://programmers.co.kr/learn/courses/30/lessons/12915
#문자열로 구성된 리스트 strings와, 정수 n이 주어졌을 때, 각 문자열의 인덱스 n번째 글자를 기준으로 오름차순 정렬하려 합니다.
# 예를 들어 strings가 [sun, bed, car]이고 n이 1이면 각 단어의 인덱스 1의 문자 u, e, a로 strings를 정렬합니다.
#나의 풀이
def solution(strings, n):
for i in range(len(strings)):
#그냥 n번째 ... |
#입력된 세 정수 a, b, c 중 가장 작은 값을 출력하는 프로그램을 작성해보자.
#(단, 조건문을 사용하지 않고 3항 연산자 ? 를 사용한다.)
#짜잔 세 개의 수를 비교하려면 삼항연산자 안에 삼항연산자를 넣어야 합니다.
#삼항연산자 이렇게 썼으니 안 잊어먹기를
#a i f (a<b) else b .. 혹시 모르니까 연습
a, b, c = input().split(" ")
a = int(a)
b = int(b)
c = int(c)
small = a if (a<b and a<c) else (b if (b<c and b<a) else c)
print(small... |
command = input('Console command: ').upper()
started = False
while command != 'QUIT':
if command != 'HELP':
print('Command not recognized.')
command = input('Console command: ').upper()
else:
print('''
Type "Start" to start the car.
Type "Stop" to stop the car.
Type "Quit" to terminate p... |
def find_max(elements):
# Redefining 'list' thru new variable
numbers = elements
# Set highest number based from 1st element in list.
high = numbers[0]
# Working Equation
for x in numbers:
if x > high:
high = x
# Value Returned to main.py
return high |
import random
# For generating any number:
print(random.random())
# Example No.01:
for x in range(10):
print(random.random())
# Example No. 02:
# For generating any number but within specific range:
for x in range(3):
print(random.randint(10, 20))
# For generating any string based from a certain list:
memb... |
a = input('What is your weight? ')
b = input("Press 'L' for lbs (pounds) or 'K' for kgs (kilograms) with respect to the unit of your inputted weight: ")
if b == 'l' or b == 'L':
c = float(a) / 2.2
print(f'You are {c} kgs.')
elif b == 'k' or b == 'K':
c = float(a) * 2.2
print(f'You are {c} lbs.')
else:
... |
""""A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.
Find the largest palindrome made from the product of two 3-digit numbers."""
listx = []
listy = []
listz = []
for x in range(100,1000):
listx.append(x)
for y in ra... |
# https://leetcode-cn.com/problems/rotate-list/submissions/
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def rotateRight(self, head: ListNode, k: int) -> ListNode:
# step 1: corner case... |
# filter(関数, イテレータ)
def is_even(n):
return n % 2 == 0
print(list(filter(is_even, range(10))))
print(list(filter(lambda n: n % 2 ==0 , range(10))))
|
# Miguel Angel Montoya
# montoya5@purdue.edu
# ID: 0031152048
# A* algorithm for the cannibals-missionaries problem
# For python 2.7
import Queue as Q
right = "Right"
left = "Left"
# Abstraction of priority queue. It lets the insertion of objects with priority.
class MyPriorityQueue(Q.PriorityQueue):
def __init... |
__author__ = 'gopalindians'
#Operators
a=3
b=3
print(a+b)
print(a-b)
print(a*b)
print(a/b)
print(a%b)
print(a**b)
print(a//b)
#Assignment operator
a+=2;
print("Now a is :" , a);
|
__author__ = 'gopalindians'
#Lists (Like arrays)
val=[1,2,3,4,5,6,7,8]
print(val[0])
#v2 using for loop
for a in val:
print(a)
#v3 using index
print(val[1:5]) #prints [2 3 4 5]
#Common operations on List
print(val) # List before appending
val.append(12)
print(val) #List after appending
|
"""
Simulates a pandemic using pygame
"""
import pygame as pg
from environment import Environment
from organisms import Organism, Health
import random, math, itertools, time, pickle
# plotting
import matplotlib.pyplot as plt
from scipy import asarray as ar,exp
import numpy as np
# size of window
(width, height) ... |
import random
import turtle
# Returns a random color!
from turtle import Turtle
def get_random_color():
return "#%06X" % (random.randint(0, 0xFFFFFF))
def get_next_color(i):
return colors[i % len(colors)]
# ====================== DO NOT EDIT THE CODE ABOVE ===========================
if __name__ == '__... |
a=[]
def even_number(x):
for i in range(0,100):
if i%2==0:
x.append(i)
print(x)
even_number(a)
|
# https://adventofcode.com/2015/day/5
import sys
import re
from pathlib import Path
path = Path(__file__).parent / "input/inputDay5.txt"
vowels = "AaEeIiOoUu"
pairs=["ab","cd","pq","xy"]
f = open(path)
def Check_Vow(string, vowels):
final = [each for each in string if each in vowels]
return fi... |
# coding: utf-8
from Tkinter import *
import time
import math
root=Tk()
root.title("画图")
running=1
#c=circle_color,d_c=dot_color
c=d_c='red'
#r=rectangle_fill_color,l_c=line_color,r_o_c=rectangle_outline_color
r=l_c=r_o_c=c_o_c='blue'
#l=line_width,d=dot_width,e=erase_width,r_o_w=rectangle_outline_width
l=d=e=r_o_w=c_o... |
# -*- coding: utf-8 -*-
# 所有名片记录的列表
cards_list = []
def show_menu():
"""显示菜单"""
print('*' * 50)
print('欢迎使用【名片管理系统】V1.0')
print('')
print('1.新建名片\n'
'2.显示全部\n'
'3.查询名片\n'
'\n'
'0.退出系统')
print('*' * 50)
def new_card():
"""新建名片"""
print('-' * 50... |
class HouseItem(object):
"""家具类"""
def __init__(self, name, area):
# 家具名称
self.name = name
# 家具占的面积
self.area = area
def __str__(self):
"""输出对象时显示的信息"""
return '家具名称:%s 占用面积:%s平米' % (self.name, self.area)
class House(object):
"""房类"""
def __init__... |
# 动态规划：将大问题分解为小问题，先解决小问题，再逐步解决大问题
# 1.当给定约束条件下优化某种指标
# 2.问题可分解为离散子问题
# 绘制表格：每个单元格都是一个子问题，将每个子问题作为一列，单元格中的值就是要优化的值，每次迭代，更新子问题中的优化值
# 背包问题
# 约束条件：背包大小4
# 优化值：背包中物品总价值
# 将4背包分解成小背包作为列
# 每次迭代小背包，更新小背包中能存放的最大价值
# 单元格中的值：
# cell[i][j] = max(cell[i-1][j], value+cell[i-1][j-当前商品重量]) # 上一单元格的值/当前商品价值+当前剩余空间的最大价值
# 注意事项：
# 1.... |
import socket
if __name__ == '__main__':
# 1. 创建服务端套接字对象
tcp_server_socket = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
# 设置端口号服用：程序退出时立即释放端口号
# SOL_SOCKET:当前套接字
# SO_REUSEADDR:复用选项
tcp_server_socket.setsockopt(socket.SOL_SOCKET, socket.SO_REUSEADDR, True)
# 2. 绑定端口号 socket.bind((... |
# 单元格中的优化值：两个字符串都包含的最长子串的长度
# 子问题：比较两个字符串的子串
# 坐标轴：两个单词
# 答案：网格中最大的数字
import numpy as np
def dynamic(str1, str2):
str_index = (0, 0)
num = 0
cell = np.zeros((len(str1), len(str2)), dtype=int);
for i in range(len(str1)):
for j in range(len(str2)):
if str1[i] == str2[j]: # 两个字母相同
if i==0 or j==0:
cel... |
class A:
a=222
def method1(self):
self.a=123
self.b=345
@classmethod
def method2(cls):
print(cls.a)
class B:
def method3(self):
self.a=999
self.b=777
s1=A()
s1.method1()
print(self.a,self.b,s1.a,s1.b)
A.method2()
... |
'''
Realiza un programa que reciba una cantidad de minutos
y muestre por pantalla a cuantas horas y minutos corresponde.
Por ejemplo: 1000 minutos son 16 horas y 40 minutos.
'''
tiempo = int(input("Introduzca minutos a convertir: "))
horas = tiempo // 60
minutos = tiempo % 60
print(tiempo, " minutos, equivalen a ", ... |
import math
'''Dados dos números, mostrar la suma, resta, división y multiplicación de ambos.'''
num1 = int(input('Introduzca primer número: '))
num2 = int(input('Introduzca segundo número: '))
print("La suma es = ",num1 + num2,
"\nLa resta es = ",num1-num2,
"\nLa multiplicación es = ",num1*num2,
"\n... |
def next_number(fibonacci_sequence):
# item -2 (penultimate) or 1
penultimate_item = fibonacci_sequence[-2] if len(fibonacci_sequence) > 1 else 0
# item -1 (last) or 1
last_item = fibonacci_sequence[-1] if bool(fibonacci_sequence) else 1
return penultimate_item + last_item
def fibonacci(n):
if not isinstan... |
#!/usr/bin/python3
hrs = input("Enter Hours:")# enter the hours
h = float(hrs)#after converting it to float
rate = input("Enter rate:")#enter the hours
r = float(rate)
def computepay(h,r):
if h <= 40:
pay = h * r
elif h > 40:
pay = 40 * r + (h-40) * r * 1.5
return pay
print(computepay(h,r))... |
import copy
def combination(arr, r):
a=[]
def generate(chosen):
if len(chosen) == r:
print(chosen)
b=copy.deepcopy(chosen)
a.append(b)
return
start = arr.index(chosen[-1]) + 1 if chosen else 0
for nxt in range(start, len(arr)):
... |
# add numbers
x = 0
def func():
b =eval(input("enter the number"))
for i in range(b+1):
if i % 2 == 0:
global x
x = i + x
print (x)
else:
print("Try next")
func()
|
class a():
def __init__(self):
self.x = 0
print ("hai i am in cons of a")
def foo(self):
print ("hai a")
class b():
def __init__(self):
self.x = 10
print ("i am in cons of b")
def foo(self):
print ("hai b")
class c(b,a):
def __init__(self):
... |
class abc():
count = 0
def __init__(self,var1 =1,var2 ="pop")
self.arg1 = var1
self.arg2 = var2
abc.count = abc.count + 1
print ("in constructor")
def print_obj(self):
print("object" , self.arg1)
print ("object", self.arg2)
def __del__(self):
... |
e = 1
som = 0
while e != 0:
e = int(input(" entrer le montant : "))
som = som + e
print(" vous devez : ", som , "euros")
|
#project_title專題題目
#name_list成員名字
name=['林郁婷','陳妤瑄','林嘉欣','甘庭銨','游佳樺','蔡柏薪','林昱承','簡郁虹','王子維']
#number_list成員學號
number=['A109260010','A109260028','A109260036','A109260038','A109260042','A109260044','A109260046','A109260074','A109260096']
#duty_list分工內容
duty=['聊天內容','程式碼']
#load_list分工比重
load=['5%','10%']
#code... |
# 1. Определение количества различных подстрок с использованием хеш-функции.
# Пусть на вход функции дана строка.
# Требуется вернуть количество различных подстрок в этой строке.
import hashlib
def subs(str):
sub = set()
for i in range(len(str)):
for j in range(len(str), i, -1):
hash_str ... |
# 1. Отсортируйте по убыванию методом пузырька одномерный целочисленный массив,
# заданный случайными числами на промежутке [-100; 100).
# Выведите на экран исходный и отсортированный массивы.
import random
SIZE = 10
MIN_ITEM = -100
MAX_ITEM = 100
array = [random.randint(MIN_ITEM, MAX_ITEM) for _ in range(SIZE)]
de... |
# 1. В диапазоне натуральных чисел от 2 до 99 определить,
# сколько из них кратны каждому из чисел в диапазоне от 2 до 9.
array = [0]*8
for i in range(2,100):
for j in range(2,10):
if i%j == 0:
array[j-2] += 1
i = 0
while i < len(array):
print(f'Числу {i+2} кратны {array[i]}')
i += 1
|
# Подсчитать, сколько было выделено памяти под переменные в ранее разработанных программах в рамках первых трех уроков.
# Проанализировать результат и определить программы с наиболее эффективным использованием памяти.
# В данной задаче массив генерируется рандомно, но присутствуют переменные, задающие его параметры.
# ... |
import pandas as pd
import matplotlib.pyplot as plt
data = { # python dictionary
'apples': [3,2,0,1],
'oranges':[0,3,7,2]
}
purchases_dataframe = pd.DataFrame(data)
print(purchases_dataframe)
dataset = pd.read_csv('data/ml-assessment-dataset.csv')
print(dataset.head(10)) # displays the header and top half of... |
"""
makes a parenthesis input valid by removing necessary characters
for example:
input: (2 + 4) * ((34 + 5)
output: (2 + 4) * (34 + 5)
"""
def makeValid(s):
paraStack = []
openStack = []
for i in range(len(s)):
if s[i] == "(" or s[i] == ")":
paraStack.append({"type":s[i], "idx... |
import tkinter
from tkinter import *
from PIL import Image, ImageTk
def capture():
window = Tk()
window.geometry("1280x800")
window.title("DSL")
window.configure(bg="#969696")
background_intro = tkinter.PhotoImage(file="Pics\capture_bckgrnd.png")
background_label = tkinter.Label... |
class Solution:
def isValid(self, s: str) -> bool:
stack = []
open_brackets = ["(", "{", "["]
closed_brackets = [")", "}", "]"]
for char in s:
if char in open_brackets:
stack.append(char)
elif char in closed_brackets:
... |
n = int(input("Enter number of columns "))
for i in range(n) :
print("*", end='')
|
word = input('Enter your word ')
characters = list(word)
from random import choice
n = len(characters)
i = 0
while i < n :
random_characters = choice(characters)
print(random_characters, end=' ')
characters.remove(random_characters)
i = i + 1
count = 0
loop = True
while loop and count < 8 :
guess = ... |
# n = int(input("n "))
# for i in range(0, n, 2):
# # range(start, stop, step)
# print(i, end=', ')
n = int(input("n "))
for i in range(n):
# i la bien gan voi cac gia tri cua n
if i % 2 == 0 :
print("Fizz")
elif i % 3 == 0 :
print("buzz")
else i % 2 % 3 == 0 :
print("Fi... |
print("The area of a circle")
r = float(input("Enter radius"))
if r < 0 :
print("The radius input must > 0")
else :
s = 3.14 * r**2
print("The area of a circle is s = {0}".format(s),"unit area")
|
'''
Class: CPSC 427
Team Member 1: Robert Brajcich
Team Member 2: Damon George
Submitted By Damon George
GU Username: dgeorge2
File Name: proj6.py
Uses breadth-first search to list states while finding goal state of given eight puzzle initial state
Usage: python proj6.py
Date: 22 February 2019
'''
from copy import de... |
INFINITY = float('inf')
#class Node():
# def __init__(self, x, y, distance):
# self.x = x
# self.y = y
# self.distance = distance
def isInsideGrid(x, y):
return 0<=x<ROW and 0<=y<COLUMN
# Moves
x = [-1, 0, 1, 0, 1, -1, 1, -1]
y = [0, 1, 0, -1, 1, 1, -1, -1]
def dijkstrasAlgo(mat, ROW, COLUMN):
# Initial... |
# To get file name
print('Enter file name')
filename = str(input())
#To check File Name
if(filename == "" or filename=="con" or filename == "aux"):
print('Error in your file name \n You cant use file names like con or aux');
else:
# To get maximum number
print("\nEnter maximum number")
... |
#Multiplication table
a = 1
while a < 11:
b = 1
while b < 11:
print('%d X %d = %d' % (a, b, (a * b)))
b += 1
a += 1 |
from Question import Question
question_prompts = [
"What's the capital of Brazil?\n(a) Rio de Janeiro\n(b) Brazilia\n(c) Sao Paolo\n\n",
"What's the capital of Australia?\n(a) Canberra\n(b) Sydney\n(c) Perth\n\n",
"What's the capital of Pakistan?\n(a) Lahore\n(b) Islamabad\n(c) Karachi\n\n"
]
qu... |
# Ecommerce Purchases Exercise
#
# In this Exercise you will be given some Fake Data about some purchases done through Amazon!
# Just go ahead and follow the directions and try your best to answer the questions and complete the tasks. Feel free to reference the solutions.
# Most of the tasks can be solved in differ... |
class Board:
def __init__(self):
self.board = self.initialize_pieces(
[[" " for i in range(8)] for j in range(8)])
self.move_list = []
def initialize_pieces(self, board) -> list:
board[1] = ['P' for i in range(8)]
board[6] = ['p' for i in range(8)]
board[0] ... |
import math
a=10
print(type(a))
print(a)
print("-------------")
b=10.9
print(type(b))
print(b)
print("-------------")
c=5+6j
print(type(c))
print(c)
print("-------------")
d=0B1010
print(type(d))
print(d)
print("-------------")
e=0XFF
print(type(e))
print(e)
print("-------------")
f=0O77
print(type(f))
print(f)
pr... |
# How to simulate animation
#required
import pygame
pygame.init()
#create a surface
gameDisplay = pygame.display.set_mode((800,600)) #initialize with a tuple
#lets add a title, aka "caption"
pygame.display.set_caption("Movement!")
pygame.display.update() #only updates portion specified
#create colors
white = (255... |
import pygame
pygame.init()
Display = pygame.display.set_mode((600,300))
pygame.display.set_caption("Bar Graph")
pygame.display.update()
white = (255,255,255)
black = (0,0,0)
red = (255, 0, 0)
green = (0, 255, 0)
yellow = (255, 255, 0)
# You will be visualizing this data.
# (Label, Color of the bar, Length of the... |
"""
다음 행렬과 같은 행렬이 있다.
m = np.array([[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14]])
1.이 행렬에서 값 7 을 인덱싱한다.
2.이 행렬에서 값 14 을 인덱싱한다.
3.이 행렬에서 배열 [6, 7] 을 슬라이싱한다.
4.이 행렬에서 배열 [7, 12] 을 슬라이싱한다.
5.이 행렬에서 배열 [[3, 4], [8, 9]] 을 슬라이싱한다.
"""
import numpy as np
a = np.array(range(15)... |
myList=[0, -5, 8, 10, 200, -100]
print(myList.index(200))
print(myList.count(0))
sortList=sorted(myList)
print(myList.reverse())
print((myList.remove(10)))
print(len(myList))
print(copyList = myList.copy())
myList.extend([4,1,3])
myList.pop()
myList.index(-5) |
num= int(input("숫자입력==>"))
if (num % 2 == 0) :
print("짝수")
else :
print("홀수") |
import numpy as np
import matplotlib.pyplot as plt
A = np.array([[1,2],[3,4],[5,6]])
print(A.shape) #(3, 2)
# 행렬의 내적
A = np.array([[1,2],[3,4]])
B = np.array([[5,6],[7,8]])
print(np.dot(A,B))
A = np.array([[1,2,3],[4,5,6]]) #2*3
B = np.array([[1,2],[3,4],[5,6]])#3*2
print(np.dot(A,B)) #2*2
# A가 2차원,B... |
money, c50000, c10000, c5000, c1000 = 0, 0, 0, 0, 0
money = int(input("지폐로 교환할 돈은 얼마?"))
c50000 = money // 50000
money %= 50000
c10000 = money // 10000
money %= 10000
c5000 = money // 5000
money %=5000
c1000 = money // 1000
money %=1000
print("\n5만원짜리==>%d장"% c50000)
print("1만원짜리==>%d장"% c10000)
print("5천원짜리==>%d... |
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