ABC291 D - Flip Cards

https://atcoder.jp/contests/abc291/tasks/abc291_d

提出

code: python

n = int(input())

ab = list(map(int, input().split())) for _ in range(n)

# pow(2, pow(10, 5))

# 両隣だけが問題

ans = 0

for i in range(n-1):

m = len(set(abi) - set(abi+1))

if m == 0:

continue

elif m == 1:

ans += 1

else:

if ans == 0:

ans = 2

else:

ans *= 2

print(ans * 2)

解答

code: python

n = int(input())

ab = list(map(int,input().split())) for _ in range(n)

mod = 998244353

# dpiflag= i枚目のカードまでの表裏の決め方のうち、条件を満たし、i枚目のカードが(flag ? 表 : 裏)であるような方法の数

dp = [0, 0 for _ in range(n)]

dp0 = 1, 1

for i in range(1, n):

for pre in range(2):

for nxt in range(2):

# 番号が異なれば

if abi-1pre != abinxt:

dpinxt += dpi-1pre % mod

print((dpn-10 + dpn-11) % mod)

テーマ

#dp

メモ

https://atcoder.jp/contests/abc291/editorial/5838

https://scrapbox.io/files/65cf23c6c80363002495ac28.png

提出

code: python

n = int(input())

ab = list(map(int, input().split())) for _ in range(n)

# print(ab)

# 1, 2], 4, 2, [3, 4

mod = 998244353

same1 = 0

same2 = 0

for i in range(n-1):

a1, b1 = abi

a2, b2 = abi+1

if (a1 == a2 and b1 == b2) or (a1 == b2 and b1 == a2):

same2 += 1

elif (a1 == a2) or (b1 == b2) or (a1 == b2) or (b1 == a2):

same1 += 1

print(same1, same2)

print(pow(2, 8) - pow(2,4)*2 - pow(2,2)*8*2)