def bubbleSort(arr):
n = len(arr)
count = 0
# Traverse through all array elements
for i in range(n-1):
# range(n) also work but outer loop will repeat one time more than needed.
# Last i elements are already in place
for j in range(0, n-i-1):
# traverse the array from 0 to n-i-1
# Swap if the element found is greater
# than the next element
if arr[j] > arr[j+1] :
arr[j], arr[j+1] = arr[j+1], arr[j]
count = count+1
return count
list = [8, 1, 7, 4, 3, 9, 2, 5, 6, 10]
count = bubbleSort(list)
print(count)
Объяснение:
ответ: 18
INPUT "A = ", A
INPUT "B = ", B
IF (999 < A)and(A < 10000)and(999 < B)and(B < 10000) THEN
IF A > B THEN
PRINT "a^3-3a^2b+3ab^2-b^3"
END IF
IF A < B THEN
PRINT "a^3+3a^2b+3ab^2+b^3"
END IF
ELSE
IF (999 > A)or(A > 10000) THEN
PRINT "A - ne ChetyrehZnachnoe"
END IF
IF (999 > B)or(B > 10000) THEN
PRINT "B - ne ChetyrehZnachnoe"
END IF
END IF
END