Information, calcul, communication
CS-119(k)
ICC-P Corrigé 4
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Exercice 1:
Exercice 2:
Exercice 3:
Exercice 4:
x: int = 10
fibo: list[int] = []
fibo.extend([0, 1])
for i in range(2, x):
fibo.append(fibo[i - 1] + fibo[i - 2])
print(fibo)
for f in fibo:
print(f)
for i, f in enumerate(fibo):
print(f"F({i}) = {f}")
Exercice 2:
from random import randint
def has_duplicates1(l: list[int]) -> bool:
for i, v in enumerate(l):
for j, w in enumerate(l):
if i != j and v == w:
return True
return False
def has_duplicates2(l: list[int]) -> bool:
length = len(l)
for i in range(length):
for j in range(i + 1, length):
if l[i] == l[j]:
return True
return False
# Bonus: Solution alternative utilisant sorted():
def has_duplicates3(l: list[int]) -> bool:
sl = sorted(l)
for i in range(len(sl)-1):
if sl[i] == sl[i+1]:
return True
return False
# Bonus: Solution alternative utilisant set():
def has_duplicates4(l: list[int]) -> bool:
# Note: set(l) converti la liste en set (ce qui retire les doublons)
return len(l) != len(set(l))
has_duplicates = has_duplicates1 # change to has_duplicates2 to test other method
print(has_duplicates([])) # False
print(has_duplicates([9, 3, 2, 5, 1])) # False
print(has_duplicates([1, 1])) # True
print(has_duplicates([1, 2, 2, 3, 1, 5, 1])) # True
def find_duplicates1(l: list[int]) -> list[int]:
dupes: list[int] = []
for i in range(len(l)):
for j in range(i + 1, len(l)):
if l[i] == l[j] and l[i] not in dupes:
dupes.append(l[i])
return dupes
def find_duplicates2(l: list[int]) -> list[int]:
dupes: list[int] = []
for i in range(len(l)):
if l[i] not in dupes:
for j in range(i + 1, len(l)):
if l[i] == l[j]:
dupes.append(l[i])
break
return dupes
# Note bonus: il est possible d'utiliser sorted ou des set là aussi...
find_duplicates = find_duplicates1 # change to find_duplicates2 to test other method
print(find_duplicates([])) # []
print(find_duplicates([1, 2, 3, 5, 9])) # []
print(find_duplicates([1, 1])) # [1]
print(find_duplicates([1, 2, 2, 3, 1, 5, 1])) # [1, 2]
def has_duplicates_alt(l: list[int]) -> bool:
return len(find_duplicates(l)) > 0
Exercice 3:
from random import randint
ints: list[int] = []
for _ in range(20):
ints.append(randint(1, 50))
print(ints)
def find_min_index(l: list[int], start_index: int = 0) -> int:
min_index = start_index
for i in range(min_index + 1, len(l)):
if l[i] < l[min_index]:
min_index = i
return min_index
print(find_min_index([3, 1, 4, 1, 18])) # 1
print(find_min_index([3, 1, 4, 1, 18], 0)) # 1, équivalent
print(find_min_index([3, 1, 4, 1, 18], start_index=2)) # 3
print(find_min_index([3, 1, 4, 1, 18], start_index=4)) # 4
def sort(l: list[int]) -> None:
for i in range(0, len(l) - 1):
j = find_min_index(l, i)
l[i], l[j] = l[j], l[i]
sort(ints)
print(ints)
Exercice 4:
# Variante 1
def binary_to_decimal(binstr: str) -> int:
acc = 0
for i, bit in enumerate(reversed(binstr)):
acc += int(bit) * 2**i
return acc
# Variante 2
def binary_to_decimal_alt(binstr: str) -> int:
acc = 0
mult = 1
for bit in reversed(binstr):
acc += int(bit) * mult
mult *= 2
return acc
# Variante 3
def binary_to_decimal_alt3(binstr: str) -> int:
acc = 0
for bit in binstr:
acc *= 2
acc += int(bit)
return acc
# Pas mal d'autres variantes possibles!
# Tests
for test_case in "101010", "10010110", "11110111110", "1000000000000000000000000000000":
print(binary_to_decimal(test_case))
print(binary_to_decimal_alt(test_case))
print(int(test_case, 2)) # version native de Python
print()