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Forwarded from Javlonbek Bo‘riyev
Forwarded from Javlonbek Bo‘riyev
Forwarded from Javlonbek Bo‘riyev
[5/8/2023 1:31 AM]
def bool_to_word(boolean):
if True:
return "No"
elif False:
return "Yes"

[5/8/2023 1:55 AM]
arr=[1,2,3,4,5,6,7,8,9,10,-11,-12,-13,-14,-15]
n=0
def count_positives_sum_negatives(arr):
for i in range(1,len(arr)):
if arr[i]>0:
return arr[i]

if arr[i]<=0:
sum(arr[i])

x=count_positives_sum_negatives(arr)
print(x)

[5/8/2023 2:02 AM]
def paperwork(n, m):
if m>0 and n>0:
return m*n
else:
return 0
Forwarded from Javlonbek Bo‘riyev
Forwarded from Javlonbek Bo‘riyev
Forwarded from Javlonbek Bo‘riyev
# massiv elementlarini 2 ga ko'paytirish # a=[1,2,3,4,8]
# def maps(a):
# for i in range(1,len(a)):
# a[i]=2*a[i]
# return a
# x=maps(a)
# print(x)
Forwarded from Javlonbek Bo‘riyev
Forwarded from Javlonbek Bo‘riyev
# You get an array of numbers, return the sum of all of the positives ones.

# # Example [1,-4,7,12] => 1 + 7 + 12 = 20

# # Note: if there is nothing to sum, the sum is default to 0.
# l=[]
# s=0
# for i in range(1,len(l)):
# if l[i]>0:
# s+=l[i]
# print(s)
# else:
# print("0")
Forwarded from Javlonbek Bo‘riyev
# Complete the square sum function so that it squares each
# number passed into it and then sums the results together.

# For example, for [1, 2, 2] it should return 9 because
# l=[]
# s=0
# for i in range(1,len(l)):
# s+=l[i]**2
# print(s)
Forwarded from Javlonbek Bo‘riyev
#Given a set of numbers, return the additive inverse of each.
#Each positive becomes negatives, and the negatives become positives.
l=[]
for i in range(0,len(l)):
l[i]=-l[i]
print(l)
Forwarded from Javlonbek Bo‘riyev
# Sinfingizda test bor edi va siz uni topshirdingiz. Tabriklaymiz!
# Ammo siz ambitsiyali odamsiz. Siz o'z sinfingizdagi o'rtacha o'quvchidan
# yaxshiroq ekanligingizni bilishni xohlaysiz.

# Siz tengdoshlaringizning test ballari bilan massiv olasiz.
# Endi o'rtachani hisoblang va balingizni solishtiring!

# Agar siz yaxshiroq bo'lsangiz, rostni qaytaring, aks holda yolg'on!

# Tarjima qilingani
b=int(input("Bahoyizni kiriting"))
l=[2,2,2,2]
a=sum(l)/len(l)
if a>b:
print("True")
else:
print("False")
Forwarded from Javlonbek Bo‘riyev
#Berilgan roʻyxatdagi raqamlarning oʻrtacha
#qiymatini hisoblaydigan funksiyani yozing.
# l=[1,2,3,4,5,6,7,89]
# s=sum(l)/len(l)
# print(s)
Forwarded from Javlonbek Bo‘riyev
# Bu kata berilgan sonni agar juft
# son bo'lsa sakkizga, aks holda to'qqizga
# ko'paytirish haqida.
# x=int(input("x="))
# if x%2==0:
# print(8*x)
# else:
# print(9*x)
Forwarded from Javlonbek Bo‘riyev
# Agar raqam 3 ga bo'linsa, "Bang",
# 5 ga bo'linsa "Boom",
# 3 va 5 ga bo'linsa "BangBoom" va "Miss" qiymatini
# (x) qabul qiladigan dastur tuzing. ularning hech
# biriga bo'linmaydi. Eslatma: Sizning dasturingiz faqat
# bitta qiymatni qaytarishi kerak
n=int(input("n="))
if n%15==0:
print("BangBoom")
elif n%5==0:
print("Bang")
else:
print("Boom")
Forwarded from Javlonbek Bo‘riyev
# Siz uydan uzoqda do'stlaringiz bilan lagerda
# # edingiz, lekin qaytish vaqti kelganida, yoqilg'i
# # tugashini va eng yaqin nasos x mil uzoqlikda ekanligini
# # tushunasiz! Bilasizmi, sizning mashinangiz o'rtacha gallon
# # uchun taxminan 25 milya ishlaydi. n gallon qoldi.

# x=int(input("Masofani kiriting="))
# n=int(input("Nechta gallon qoldi="))
# if n*25>=x:
# print("Yetadi")
# else:
# print("Yetmaydi")
Forwarded from Javlonbek Bo‘riyev
# Funktsiyani to'ldiring, shunda u o'ziga berilgan uchta
# ballning o'rtacha qiymatini topadi va shu baho bilan bog'liq
# harf qiymatini qaytaradi.


# Numerical Score Letter Grade
# 90 <= score <= 100 'A'
# 80 <= score < 90 'B'
# 70 <= score < 80 'C'
# 60 <= score < 70 'D'
# 0 <= score < 60 'F'
def get_grade(s1, s2, s3):
if 0<=s1<60:
return "F"
if 60<=s1<70:
return "F"
if 70<=s1<80:
return "F"
if 80<=s1<90:
return "F"
if 90<=s1<100:
return "F"
Forwarded from Javlonbek Bo‘riyev
# Berilgan massivning barcha raqamlarini yig'ing
# # ( cq. list ), eng yuqori va eng past elementdan tashqari
# # (indeks bo'yicha emas, qiymat bo'yicha! ).

# # Bir xil qiymatga ega bo'lgan bir nechta element bo'lsa
# # ham, mos ravishda eng yuqori yoki eng past element har
# # bir chekkada bitta elementdir.
# l=[1,2,3,4,5,6,7,8,9]
# s=sum(l)-(max(l)+min(l))
# print(s)
Forwarded from Javlonbek Bo‘riyev
# Qahramon o'z vazifasini bajarish uchun qasrga ketmoqda.
# Biroq, unga qasr bir nechta kuchli ajdarlar bilan
# o'ralganligini aytishdi! Har bir ajdaho mag'lub bo'lish
# uchun 2 ta o'q oladi, bizning qahramonimiz qancha o'q olib
# yurishi kerakligini bilmaydi. Agar u ma'lum miqdordagi o'qni
# ushlab, boshqa ma'lum miqdordagi ajdarlar bilan jang qilish
# uchun oldinga siljiydi deb faraz qilsak, u omon qoladimi?

# Ha bo'lsa rost, aks holda noto'g'ri deb qaytaring :)
x=int(input("ajdarlar soni="))
n=int(input("oqlar son="))
if n>=2**x:
print("rost")
else:
print("Yolg'on")
Forwarded from Javlonbek Bo‘riyev
# Massivni qaytaring, bunda birinchi element musbat sonlar soni,
# ikkinchi element esa manfiy sonlar yig‘indisidir.

l=[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, -11, -12, -13, -14, -15]
# , siz [10, -65] ni qaytarishingiz kerak.

n=0
s=0
for i in range(len(l)):
if l[i]>0:
n+=1

else:
s+=l[i]

print(n,"musbat elementlari soni")
print(s,"manfiy elementlari yig'indisi")
Forwarded from Javlonbek Bo‘riyev
# Ta'tilda sayohat qilish uchun sizga ijara mashinasi kerak bo'ladi.
# \Avtomobil ijarasi menejeri sizga yaxshi takliflar beradi.

# Har kuni mashina ijarasi 40 dollar turadi.
# Agar siz mashinani 7 yoki undan ortiq kunga ijaraga olsangiz,
# jami 50 dollar chegirma olasiz. Shu bilan bir qatorda, agar siz
# mashinani 3 yoki undan ortiq kunga ijaraga olsangiz, jami 20 AQSh
# dollari miqdorida chegirma olasiz.

# Turli kunlar uchun umumiy miqdorni ko'rsatadigan kodni yozing(d).
n=int(input("car="))
d=int(input("day="))
if d<7 and n<3:
s=40*d*n
print("Jami:",s)
if d>7 and n<3:
s=40*d*n-50
print("Jami:",s)
if d>7 and n>3:
s=40*d*n-70
print("Jami:",s)
if d<7 and n>3:
s=40*d*n-20
print("Jami:",s)