maximize the number of horizontal bricks <=> use as many 1*2 (smallest) bricks as possible
m is even => (horizontal, vertical) = (n * m // 2, 0)
m is odd => (horizontal, vertical) = (n * (m - 1) // 2, 0), for the last 3 columns, we use 1*3 bricks so that we don't need to use vertical bricks.
for _ in range(int(input())):
n, m = map(int, input().split())
if m % 2 == 0:
print(n * m // 2)
else:
print(n * (m - 1) // 2)def solve(n: int, a: list[int]) -> tuple[int, list[tuple[int, int]]]:
if 0 in a:
return 0, []
if sum(1 for x in a if x < 0) % 2 == 0:
return 1, [(1, 0)]
return 0, []
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
k, ops = solve(n, a)
print(k)
for i in range(k):
print(f"{ops[i][0]} {ops[i][1]}")Accordingly, the string doesn't match the template if the condition doesn't hold for at least one i.
To make the string c does not match the template, we only need one position that does not meet the criteria.
When c[i] equals to a[i] or b[i], we cannot create a template that a[i] and b[i] match and c[i] does not.
a[i] = “a”, b[i] = “b”, c[i] = “c” => t[i] = “C”
a[i] = “a”, b[i] = “b”, c[i] = “a” => cannot make a[i] meet the criteria and c[i] does not as they are the same.
def exists(n: int, a: str, b: str, c: str) -> bool:
for i in range(n):
if c[i] != a[i] and c[i] != b[i]:
return True
return False
for _ in range(int(input())):
n = int(input())
a = input()
b = input()
c = input()
if exists(n, a, b, c):
print("YES")
else:
print("NO")Since tracking calories intake did not work this time and I'm getting overweight, let's try low carbohydrate diet for a few months.
MCT oil is not sold at the nearest Safeway. I might want to check Target tomorrow.
Eggs 20g salad 20g Sashimi 5g Meat stick 0g
Total carbohydrate 45g (< 60g)
TODO: