[PYTHON] Remplir au codeur

introduction

Ceci est un journal pour les débutants en programmation compétitive pour résoudre les problèmes d'atcoder. L'objectif principal est d'encourager la poursuite en publiant le journal. Puisqu'il existe des paramètres et des explications de problème officiels, je les omettrai. Je posterai les impressions (commentaires) que j'ai résolus, les sites et codes auxquels j'ai fait référence dans le revêtement. De plus, puisqu'il restera comme journal de débutant, les codes autres que la bonne réponse seront également affichés. Plus précisément, je publierai un code qui correspond à la politique mais devient TLE comme référence.

ABC

Problème C

147 - HonestOrUnkind2

Code 1. recherche complète de bits

time: O(2^N) space: O(N)

def solve(N, A, X, Y):
    ret = 0
    for i in range(2 ** N):
        tmp = []
        for j in range(N):
            if i >> j & 1:
                tmp.append(j)
        if not tmp:
            continue

        is_valid = True
        for a_idx in tmp:
            for idx in range(A[a_idx]):
                if (Y[a_idx][idx] and X[a_idx][idx] - 1 not in tmp) \
                        or (not Y[a_idx][idx] and X[a_idx][idx] - 1 in tmp):
                    is_valid = False
                    break
        ret = max(ret, len(tmp)) if is_valid else ret
    return ret

if __name__ == "__main__":
    N = int(input())
    A = []
    X, Y = [], []
    for _ in range(N):
        A.append(int(input()))
        tmp_x, tmp_y = [], []
        for _ in range(A[-1]):
            x, y = map(int, input().split())
            tmp_x.append(x)
            tmp_y.append(y)
        X.append(tmp_x)
        Y.append(tmp_y)
    print(solve(N, A, X, Y))

165 - Many Requirements

Code 1. DFS

time: O(Q_{M}H_{N})

def solve(cur_index):
    global ret
    if cur_index == N:
        tmp_sum = 0
        for i in range(Q):
            if A[b[i]] - A[a[i]] == c[i]:
                tmp_sum += d[i]
        ret = max(ret, tmp_sum)
        return
    for i in range(min(A[cur_index], M), M + 1):
        A[cur_index + 1] = i
        solve(cur_index + 1)


if __name__ == '__main__':
    N, M, Q = map(int, input().split())
    a = [0 for _ in range(Q)]
    b = [0 for _ in range(Q)]
    c = [0 for _ in range(Q)]
    d = [0 for _ in range(Q)]
    for i in range(Q):
         a[i], b[i], c[i], d[i] = map(int, input().split())
    ret = 0
    A = [1 for _ in range(N + 1)]
    solve(0)
    print(ret)

167 - Skill Up

Code 1. recherche complète de bits

time: O(M2^N) space: O(N) + O(M)

def is_valid(vals, M, X):
    for i in range(M):
        if vals[i] < X:
            return False
    return True

def solve(N, M, X, C, A):
    ret = float("inf")
    for i in range(2 ** N):
        tmp_vals = [0 for _ in range(M)]
        cost = 0
        for j in range(N):
            if (i >> j) & 1:
                cost += C[j]
                for k in range(M):
                   tmp_vals[k] += A[j][k]
        ret = min(ret, cost) if is_valid(tmp_vals, M, X) else ret
    return ret if ret != float("inf") else -1

if __name__ == '__main__':
    N, M, X = map(int, input().split())
    C = []
    A = []
    for _ in range(N):
        tmp = list(map(int, input().split()))
        C.append(tmp[0])
        A.append(tmp[1:])

    print(solve(N, M, X, C, A))

Problème D

145 - Knight

Je pensais quand j'ai commencé à résoudre des choses comme "Puis-je faire des coordonnées obliques ...?", Mais c'est impossible. En passant, l'accélération du coefficient binomial semble être un paramètre courant.

Code 1. Mise en œuvre naïve

time: O\ (X + Y) Une fonction spéciale sur la façon de trouver "trop divisé par 1000000007"! ~ De l'élément inverse au logarithme discret ~

X, Y = map(int, input().split())
n = (2 * X - Y) / 3
m = (2 * Y - X) / 3

if n < 0 or m < 0 or not n.is_integer() or not m.is_integer():
    print(0)
    exit()

n, m = map(int, [n, m])
mod = 10**9 + 7
fact = [1]
inv = [1]
for i in range(1, X + Y + 1):
    fact.append((fact[i-1] * i) % mod)
    inv.append(pow(fact[-1], mod-2, mod))

def cmb(n, r):
    return fact[n] * inv[n-r] * inv[r]

print(cmb(n+m, min(n, m)) % mod)

Code 2. Introduction de ModInt + Accélération du calcul d'élément inverse

Lors de la soumission, il est préférable de ne décrire que la méthode nécessaire. time: O\ (X + Y) J'ai implémenté modint en Python

X, Y = map(int, input().split())
n = (2 * X - Y) / 3
m = (2 * Y - X) / 3

if n < 0 or m < 0 or not n.is_integer() or not m.is_integer():
    print(0)
    exit()

MOD = 10**9 + 7
class ModInt:
    def __init__(self, x):
        self.x = x % MOD

    def __str__(self):
        return str(self.x)

    __repr__ = __str__

    def __add__(self, other):
        return (
            ModInt(self.x + other.x) if isinstance(other, ModInt) else
            ModInt(self.x + other)
        )

    def __sub__(self, other):
        return (
            ModInt(self.x - other.x) if isinstance(other, ModInt) else
            ModInt(self.x - other)
        )

    def __mul__(self, other):
        return (
            ModInt(self.x * other.x) if isinstance(other, ModInt) else
            ModInt(self.x * other)
        )

    def __truediv__(self, other):
        return (
            ModInt(
                self.x * pow(other.x, MOD - 2, MOD)
            ) if isinstance(other, ModInt) else
            ModInt(self.x * pow(other, MOD - 2, MOD))
        )

    def __pow__(self, other):
        return (
            ModInt(pow(self.x, other.x, MOD)) if isinstance(other, ModInt) else
            ModInt(pow(self.x, other, MOD))
        )

    __radd__ = __add__

    def __rsub__(self, other):
        return (
            ModInt(other.x - self.x) if isinstance(other, ModInt) else
            ModInt(other - self.x)
        )

    __rmul__ = __mul__

    def __rtruediv__(self, other):
        return (
            ModInt(
                other.x * pow(self.x, MOD - 2, MOD)
            ) if isinstance(other, ModInt) else
            ModInt(other * pow(self.x, MOD - 2, MOD))
        )

    def __rpow__(self, other):
        return (
            ModInt(pow(other.x, self.x, MOD)) if isinstance(other, ModInt) else
            ModInt(pow(other, self.x, MOD))
        )

n, m = map(int, [n, m])
fact = [ModInt(1)]
for i in range(1, X + Y + 1):
    fact.append(fact[i-1] * i)

def cmb(n, r):
    return fact[n] / fact[n-r] / fact[r]

print(cmb(n+m, min(n, m)))

146 - Coloring Edges on Tree

Code 1. Super Stupid BFS (TLE)

Eh bien, l'idée elle-même est correcte. Malheureusement, cela n'a pas assez de temps de calcul. Le fait est qu'en faisant BFS, je cherche à l'intérieur de ʻused_edges et ʻused_colors avec la fonction ʻin`. Les ordres de calcul augmentent ici. La bonne réponse peut être obtenue en éliminant ce calcul supplémentaire. time: O(N^2)


from collections import deque
N = int(input())
edges = [[] for _ in range(N + 1)]
for edge_idx in range(1, N):
    a, b = map(int, input().split())
    edges[a].append([edge_idx, b])
    edges[b].append([edge_idx, a])

used_edges = []
colors = [1 for _ in range(1, N + 1)]
used_colors = [[] for _ in range(N + 1)]
stack = deque([[edges[1], 0]])
max_col = 1

while stack:
    edge_info, parent_node_idx = stack.popleft()
    cur_color = 1
    for edge_idx, node_idx in edge_info:
        if edge_idx in used_edges: continue
        while cur_color in used_colors[node_idx] or cur_color in used_colors[parent_node_idx]:
            cur_color += 1
        used_colors[node_idx].append(cur_color)
        colors[edge_idx] = cur_color
        stack.append([edges[node_idx], node_idx])
        used_edges.append(edge_idx)
        max_col = max(max_col, cur_color)
        cur_color += 1

print(max_col)
for c in colors[1:]:
    print(c)

Code 2. BFS amélioré

Changement de la structure de données utilisée dans ʻused_edges et ʻused_colors from list en set. Cela a changé l'ordre moyen des recherches ʻin` de $ O (n) $ à $ O (1) $. Time Complexity - Python Wiki time: O(N)

from collections import deque
N = int(input())
edges = [[] for _ in range(N + 1)]
for edge_idx in range(1, N):
    a, b = map(int, input().split())
    edges[a].append([edge_idx, b])
    edges[b].append([edge_idx, a])

used_edges = set()
colors = [1 for _ in range(1, N + 1)]
used_colors = [set() for _ in range(N + 1)]
stack = deque([[edges[1], 0]])
max_col = 1

while stack:
    edge_info, parent_node_idx = stack.pop()
    cur_color = 1
    for edge_idx, node_idx in edge_info:
        if edge_idx in used_edges: continue
        while cur_color in used_colors[node_idx] or cur_color in used_colors[parent_node_idx]:
            cur_color += 1
        used_colors[node_idx].add(cur_color)
        colors[edge_idx] = cur_color
        stack.append([edges[node_idx], node_idx])
        used_edges.add(edge_idx)
        max_col = max(max_col, cur_color)
        cur_color += 1

print(max_col)
for c in colors[1:]:
    print(c)

147 - Xor Sum 4

Code 1. Double boucle (TLE)

Implémentation utilisant le "ModInt" mentionné ci-dessus. Eh bien, c'est une solution que vous voulez absolument essayer au cas où, même si elle ne passe jamais en termes de montant de calcul. Inutile de dire que tourner la double boucle est TLE. time: O(N^2)

MOD = 10**9 + 7
class ModInt:
    def __init__(self, x):
        if isinstance(x, str):
            x = int(x)
        self.x = x % MOD

    def __str__(self):
        return str(self.x)

    __repr__ = __str__

    def __add__(self, other):
        return (
            ModInt(self.x + other.x) if isinstance(other, ModInt) else
            ModInt(self.x + other)
        )

    def __sub__(self, other):
        return (
            ModInt(self.x - other.x) if isinstance(other, ModInt) else
            ModInt(self.x - other)
        )

    def __mul__(self, other):
        return (
            ModInt(self.x * other.x) if isinstance(other, ModInt) else
            ModInt(self.x * other)
        )

    def __truediv__(self, other):
        return (
            ModInt(
                self.x * pow(other.x, MOD - 2, MOD)
            ) if isinstance(other, ModInt) else
            ModInt(self.x * pow(other, MOD - 2, MOD))
        )

    def __pow__(self, other):
        return (
            ModInt(pow(self.x, other.x, MOD)) if isinstance(other, ModInt) else
            ModInt(pow(self.x, other, MOD))
        )

    __radd__ = __add__

    def __rsub__(self, other):
        return (
            ModInt(other.x - self.x) if isinstance(other, ModInt) else
            ModInt(other - self.x)
        )

    __rmul__ = __mul__

    def __rtruediv__(self, other):
        return (
            ModInt(
                other.x * pow(self.x, MOD - 2, MOD)
            ) if isinstance(other, ModInt) else
            ModInt(other * pow(self.x, MOD - 2, MOD))
        )

    def __rpow__(self, other):
        return (
            ModInt(pow(other.x, self.x, MOD)) if isinstance(other, ModInt) else
            ModInt(pow(other, self.x, MOD))
        )

N = int(input())
A = list(map(int, input().split()))

def xor(a, b):
    return a ^ b

ret = ModInt(0)
for i in range(N - 1):
    for j in range(i + 1, N):
        ret += xor(A[i], A[j])

print(ret)
Code 2. Calcul XOR (TLE) pour chaque bit

C'est une solution hypothétique, mais c'est TLE. time: O(Nlog(max(A)))

MOD = 10**9 + 7
class ModInt:
    def __init__(self, x):
        if isinstance(x, str):
            x = int(x)
        self.x = x % MOD

    def __str__(self):
        return str(self.x)

    __repr__ = __str__

    def __add__(self, other):
        return (
            ModInt(self.x + other.x) if isinstance(other, ModInt) else
            ModInt(self.x + other)
        )

    def __sub__(self, other):
        return (
            ModInt(self.x - other.x) if isinstance(other, ModInt) else
            ModInt(self.x - other)
        )

    def __mul__(self, other):
        return (
            ModInt(self.x * other.x) if isinstance(other, ModInt) else
            ModInt(self.x * other)
        )

    def __truediv__(self, other):
        return (
            ModInt(
                self.x * pow(other.x, MOD - 2, MOD)
            ) if isinstance(other, ModInt) else
            ModInt(self.x * pow(other, MOD - 2, MOD))
        )

    def __pow__(self, other):
        return (
            ModInt(pow(self.x, other.x, MOD)) if isinstance(other, ModInt) else
            ModInt(pow(self.x, other, MOD))
        )

    __radd__ = __add__

    def __rsub__(self, other):
        return (
            ModInt(other.x - self.x) if isinstance(other, ModInt) else
            ModInt(other - self.x)
        )

    __rmul__ = __mul__

    def __rtruediv__(self, other):
        return (
            ModInt(
                other.x * pow(self.x, MOD - 2, MOD)
            ) if isinstance(other, ModInt) else
            ModInt(other * pow(self.x, MOD - 2, MOD))
        )

    def __rpow__(self, other):
        return (
            ModInt(pow(other.x, self.x, MOD)) if isinstance(other, ModInt) else
            ModInt(pow(other, self.x, MOD))
        )

def encode(N, A, max_k):
    bits = [[0, 0] for _ in range(max_k)]
    for i in range(N):
        a = format(A[i], 'b').zfill(max_k)
        for k in range(max_k):
            if int(a[k]) == 0:
                bits[max_k - 1 - k][0] += 1
            else:
                bits[max_k - 1 - k][1] += 1
    return bits

def decode(bits):
    return sum([ModInt(2 ** k * (counter[0] * counter[1])) for k, counter in enumerate(bits)])

N = int(input())
A = list(map(int, input().split()))
max_k = len(bin(max(A))) - 2

print(decode(encode(N, A, max_k)))

Code 3. Calcul XOR pour chaque bit (accélération)

Vous pouvez accélérer en faisant cela. C'est presque quatre fois plus rapide. time: O(Nlog(max(A)))

MOD = 10**9 + 7
class ModInt:
    def __init__(self, x):
        if isinstance(x, str):
            x = int(x)
        self.x = x % MOD

    def __str__(self):
        return str(self.x)

    __repr__ = __str__

    def __add__(self, other):
        return (
            ModInt(self.x + other.x) if isinstance(other, ModInt) else
            ModInt(self.x + other)
        )

    def __sub__(self, other):
        return (
            ModInt(self.x - other.x) if isinstance(other, ModInt) else
            ModInt(self.x - other)
        )

    def __mul__(self, other):
        return (
            ModInt(self.x * other.x) if isinstance(other, ModInt) else
            ModInt(self.x * other)
        )

    def __truediv__(self, other):
        return (
            ModInt(
                self.x * pow(other.x, MOD - 2, MOD)
            ) if isinstance(other, ModInt) else
            ModInt(self.x * pow(other, MOD - 2, MOD))
        )

    def __pow__(self, other):
        return (
            ModInt(pow(self.x, other.x, MOD)) if isinstance(other, ModInt) else
            ModInt(pow(self.x, other, MOD))
        )

    __radd__ = __add__

    def __rsub__(self, other):
        return (
            ModInt(other.x - self.x) if isinstance(other, ModInt) else
            ModInt(other - self.x)
        )

    __rmul__ = __mul__

    def __rtruediv__(self, other):
        return (
            ModInt(
                other.x * pow(self.x, MOD - 2, MOD)
            ) if isinstance(other, ModInt) else
            ModInt(other * pow(self.x, MOD - 2, MOD))
        )

    def __rpow__(self, other):
        return (
            ModInt(pow(other.x, self.x, MOD)) if isinstance(other, ModInt) else
            ModInt(pow(other, self.x, MOD))
        )

def calc(N, A, max_k):
    ret = ModInt(0)
    for k in range(max_k):
        count = 0
        for i in A:
            if i & (1 << k):
                count += 1
        ret += 2 ** k * count * (N - count)
    return ret

N = int(input())
A = list(map(int, input().split()))
max_k = 60

print(calc(N, A, max_k))

148 - Brick Break

Code 1. Italate

Je n'ai rien à dire. Quand j'ai lu le sens, j'avais un peu peur que ce soit trop simple et j'ai oublié quelque chose. time: O(N)


N = int(input())
a = list(map(int, input().split()))
broken = 0

for n in range(N):
    if a[n] != broken + 1:
        continue
    broken += 1

if broken == 0:
    print(-1)
else:
    print(N - broken)

149 - Prediction and Restriction

Code 1. Italate

time: O(N)


N, K = map(int, input().split())
R, S, P = map(int, input().split())
T = input()

def getScore(t, R, S, P):
    if t == 'r':
        return P
    if t == 's':
        return R
    return S

latest = [{'string': '', 'count': 0} for _ in range(K)]

ret = 0
for idx, t in enumerate(T):
    latest_idx = idx % K
    if idx + 1 <= K or t != latest[latest_idx]['string']:
        ret += getScore(t, R, S, P)
        latest[latest_idx]['string'] = t
        latest[latest_idx]['count'] = 1
        continue

    if latest[latest_idx]['count'] % 2 == 0:
        ret += getScore(t, R, S, P)
    latest[latest_idx]['count'] += 1

print(ret)

150 - Semi Common Multiple

Code 1. Italate

J'ai eu du mal parce que je l'ai négligé dans ma considération. AtCoder ABC 150 D - Semi-commun multiple (400 points) time: O(N)

from fractions import gcd

def lcm(x, y):
    return (x * y) // gcd(x, y)

def solve(N, M, A):
    while A[0] % 2 == 0:
        for idx, a in enumerate(A):
            if a % 2 != 0:
                return 0
            A[idx] //= 2
        M //= 2

    for a in A:
        if a % 2 == 0:
            return 0

    cur_lcm = 1
    for a in A:
        cur_lcm = lcm(cur_lcm, a)
        if M < cur_lcm:
            return 0
    return (M // cur_lcm + 1) // 2

if __name__ == '__main__':
    N, M = map(int, input().split())
    A = list((map(int, input().split())))
    A = [a // 2 for a in A]
    print(solve(N, M, A))

151 - Maze Master

Code 1. BFS + double boucle

Mettre à jour ʻused après stack.popleft () ʻau lieu de stack.append est TLE.

time: O((HW)^2)

from collections import deque

def bfs(start, maze, H, W):
    max_step = 0
    used = [[False for _ in range(W)] for _ in range(H)]
    start_x, start_y = start
    used[start_x][start_y] = True
    stack = deque([[start, 0]])
    while stack:
        (i, j), step = stack.popleft()
        if maze[i][j] == '#':
            continue
        max_step = max(max_step, step)
        for x, y in [(i + 1, j), (i - 1, j), (i, j + 1), (i, j - 1)]:
            if 0 <= x < H and 0 <= y < W and not used[x][y]:
                stack.append([(x, y), step + 1])
                used[x][y] = True
    return max_step

if __name__ == '__main__':
    H, W = map(int, input().split())
    S = []
    for _ in range(H):
        S.append(list(input()))
    ret = 0
    for i in range(H):
        for j in range(W):
            ret = max(ret, bfs((i, j), S, H, W))
    print(ret)

152 - Handstand 2

Code 1. génération de carte

Remarquez-vous que vous pouvez stocker tous les nombres dans un tableau bidimensionnel de longueur constante? time: O(N)

def solve(N):
    ret = 0
    counter = [[0 for _ in range(10)] for _ in range(10)]
    for n in range(1, N + 1):
        i, j = map(int, [str(n)[0], str(n)[-1]])
        counter[i][j] += 1

    for i in range(10):
        for j in range(10):
            ret += counter[i][j] * counter[j][i]
    return ret


if __name__ == '__main__':
    N = int(input())
    ret = solve(N)
    print(ret)

156 - Bouquet

Code 1. Mettre en italique et multiplier (TLE)

Au fait, je ne sais pas comment calculer le multiplicateur plus rapidement que l'ordre linéaire. time: O(N)

MOD = 10**9 + 7
class ModInt:
    def __init__(self, x):
        self.x = x % MOD

    def __str__(self):
        return str(self.x)

    __repr__ = __str__

    def __add__(self, other):
        return (
            ModInt(self.x + other.x) if isinstance(other, ModInt) else
            ModInt(self.x + other)
        )

    def __sub__(self, other):
        return (
            ModInt(self.x - other.x) if isinstance(other, ModInt) else
            ModInt(self.x - other)
        )

    def __mul__(self, other):
        return (
            ModInt(self.x * other.x) if isinstance(other, ModInt) else
            ModInt(self.x * other)
        )

    def __truediv__(self, other):
        return (
            ModInt(
                self.x * pow(other.x, MOD - 2, MOD)
            ) if isinstance(other, ModInt) else
            ModInt(self.x * pow(other, MOD - 2, MOD))
        )

    def __pow__(self, other):
        return (
            ModInt(pow(self.x, other.x, MOD)) if isinstance(other, ModInt) else
            ModInt(pow(self.x, other, MOD))
        )

    __radd__ = __add__

    def __rsub__(self, other):
        return (
            ModInt(other.x - self.x) if isinstance(other, ModInt) else
            ModInt(other - self.x)
        )

    __rmul__ = __mul__

    def __rtruediv__(self, other):
        return (
            ModInt(
                other.x * pow(self.x, MOD - 2, MOD)
            ) if isinstance(other, ModInt) else
            ModInt(other * pow(self.x, MOD - 2, MOD))
        )

    def __rpow__(self, other):
        return (
            ModInt(pow(other.x, self.x, MOD)) if isinstance(other, ModInt) else
            ModInt(pow(other, self.x, MOD))
        )

def cmb(fact, n, r):
    return fact[n] / fact[n-r] / fact[r]

def solve(n, a, b):
    fact = [ModInt(1)]
    for i in range(1, max(n, a, b) + 1):
        fact.append(fact[i-1] * i)

    return  ModInt(2**n) - 1 - cmb(fact, n, a) - cmb(fact, n, b)

if __name__ == '__main__':
    n, a, b = map(int, input().split())
    ret = solve(n, a, b)
    print(ret)
Code 2. Calcul du nCr amélioré

Pour une raison quelconque, j'ai mal compris que l'itération de $ n-r + 1 $ ~ $ n $ était $ O (N) $ (pourquoi vraiment?). J'ai essayé d'implémenter la méthode square à plusieurs reprises, mais il semble que ce n'est pas nécessaire en python car elle est déjà adoptée dans la fonction pow.

time: max(O(A), O(B))

MOD = 10**9 + 7
class ModInt:
    def __init__(self, x):
        self.x = x % MOD

    def __str__(self):
        return str(self.x)

    __repr__ = __str__

    def __add__(self, other):
        return (
            ModInt(self.x + other.x) if isinstance(other, ModInt) else
            ModInt(self.x + other)
        )

    def __sub__(self, other):
        return (
            ModInt(self.x - other.x) if isinstance(other, ModInt) else
            ModInt(self.x - other)
        )

    def __mul__(self, other):
        return (
            ModInt(self.x * other.x) if isinstance(other, ModInt) else
            ModInt(self.x * other)
        )

    def __truediv__(self, other):
        return (
            ModInt(
                self.x * pow(other.x, MOD - 2, MOD)
            ) if isinstance(other, ModInt) else
            ModInt(self.x * pow(other, MOD - 2, MOD))
        )

    def __pow__(self, other):
        return (
            ModInt(pow(self.x, other.x, MOD)) if isinstance(other, ModInt) else
            ModInt(pow(self.x, other, MOD))
        )

    __radd__ = __add__

    def __rsub__(self, other):
        return (
            ModInt(other.x - self.x) if isinstance(other, ModInt) else
            ModInt(other - self.x)
        )

    __rmul__ = __mul__

    def __rtruediv__(self, other):
        return (
            ModInt(
                other.x * pow(self.x, MOD - 2, MOD)
            ) if isinstance(other, ModInt) else
            ModInt(other * pow(self.x, MOD - 2, MOD))
        )

    def __rpow__(self, other):
        return (
            ModInt(pow(other.x, self.x, MOD)) if isinstance(other, ModInt) else
            ModInt(pow(other, self.x, MOD))
        )

def cmb(fact, n, r):
    mul = ModInt(1)
    for i in range(n - r + 1, n + 1):
        mul *= i
    return mul / fact[r]

def solve(n, a, b):
    fact = [ModInt(1)]
    for i in range(1, max(a, b) + 1):
        fact.append(fact[i-1] * i)
    return  ModInt(pow(2,n)) - 1 - cmb(fact, n, a) - cmb(fact, n, b)

if __name__ == '__main__':
    n, a, b = map(int, input().split())
    ret = solve(n, a, b)
    print(ret)

157 - Friend Suggestions

Code 1. Stupid Union Find (TLE)

time: O(N^2)

class UnionFind():
    def __init__(self, n):
        self.n = n
        self.parents = [-1] * n

    def find(self, x):
        if self.parents[x] < 0:
            return x

        self.parents[x] = self.find(self.parents[x])
        return self.parents[x]

    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)

        if x == y:
            return

        if self.parents[x] > self.parents[y]:
            x, y = y, x

        self.parents[x] += self.parents[y]
        self.parents[y] = x

    def members(self, x):
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]

def solve(uf, first_order_friends, blocked_list):
    ret = [0]
    for i in range(1, N + 1):
        ret.append(len(list(set(uf.members(i)) - first_order_friends[i] - blocked_list[i])) - 1)
    return ret

if __name__ == '__main__':
    N, M, K = map(int, input().split())
    first_order_friends = [set() for _ in range(N + 1)]
    blocked_list = [set() for _ in range(N + 1)]
    uf = UnionFind(N + 1)

    for _ in range(M):
        a, b = map(int, input().split())
        first_order_friends[a].add(b)
        first_order_friends[b].add(a)
        uf.union(a, b)

    for _ in range(K):
        c, d = map(int, input().split())
        blocked_list[c].add(d)
        blocked_list[d].add(c)

    for ans in solve(uf, first_order_friends, blocked_list)[1:]:
        print(ans, end=' ')
Code 2. Double boucle + Union Find (TLE)

Unifiés first_order_friends et blocks_list. À ce stade, j'ai réalisé que tourner la double boucle était la cause directe de TLE. time: O(N^2)

class UnionFind():
    def __init__(self, n):
        self.n = n
        self.parents = [-1] * n

    def find(self, x):
        if self.parents[x] < 0:
            return x

        self.parents[x] = self.find(self.parents[x])
        return self.parents[x]

    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)

        if x == y:
            return

        if self.parents[x] > self.parents[y]:
            x, y = y, x

        self.parents[x] += self.parents[y]
        self.parents[y] = x

    def members(self, x, except_list):
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root and i not in except_list]

def solve(uf, first_order_friends):
    ret = [0]
    for i in range(1, N + 1):
        ret.append(len(uf.members(i, first_order_friends[i])) - 1)
    return ret

if __name__ == '__main__':
    N, M, K = map(int, input().split())
    first_order_friends = [set() for _ in range(N + 1)]
    blocked_list = [set() for _ in range(N + 1)]
    uf = UnionFind(N + 1)

    for _ in range(M):
        a, b = map(int, input().split())
        first_order_friends[a].add(b)
        first_order_friends[b].add(a)
        uf.union(a, b)

    for _ in range(K):
        c, d = map(int, input().split())
        first_order_friends[c].add(d)
        first_order_friends[d].add(c)

    for ans in solve(uf, first_order_friends)[1:]:
        print(ans, end=' ')

Code 3. Contrôle d'entrée + UnionFind

Basé sur la considération dans le code 2

--Ajouter à ʻexcept_list` seulement si c et d appartiennent au même groupe -Obtenir le nombre d'éléments dans le groupe avec $ O (1) $

Je l'ai changé en.

time: O(N)

class UnionFind():
    def __init__(self, n):
        self.n = n
        self.parents = [-1] * n

    def find(self, x):
        if self.parents[x] < 0:
            return x

        self.parents[x] = self.find(self.parents[x])
        return self.parents[x]

    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)

        if x == y:
            return

        if self.parents[x] > self.parents[y]:
            x, y = y, x

        self.parents[x] += self.parents[y]
        self.parents[y] = x

    def same(self, x, y):
        return self.find(x) == self.find(y)

    def size(self, x):
        return -self.parents[self.find(x)]

def solve(uf, except_list):
    ret = [0]
    for i in range(1, N + 1):
        ret.append(uf.size(i) - 1 - len(except_list[i]))
    return ret

if __name__ == '__main__':
    N, M, K = map(int, input().split())
    except_list = [[] for _ in range(N + 1)]
    uf = UnionFind(N + 1)

    for _ in range(M):
        a, b = map(int, input().split())
        except_list[a].append(b)
        except_list[b].append(a)
        uf.union(a, b)

    for _ in range(K):
        c, d = map(int, input().split())
        if not uf.same(c, d):
            continue
        except_list[c].append(d)
        except_list[d].append(c)

    for ans in solve(uf, except_list)[1:]:
        print(ans, end=' ')

160 - Line++

Code 1. BFS

J'avais peur que 10 $ ^ 6 $ suffiraient, mais cela semble bien. time: O(N^2)

from collections import deque

def solve(N, X, Y):
    ret = [0] * N
    already_used = [[False for _ in range(N + 1)] for _ in range(N + 1)]
    for i in range(1, N + 1):
        already_used[i][i] = True
        reached = [False for _ in range(N + 1)]
        reached[i] = True
        stack = deque([[i, 0]])

        while stack:
            node, step = stack.popleft()
            if 1 <= step and not already_used[i][node]:
               ret[step] += 1
            already_used[i][node], already_used[node][i] = True, True

            for x in [node + 1, node - 1]:
                if 0 < x < N + 1 and not reached[x]:
                    stack.append([x, step + 1])
                    reached[x] = True
            if node == X and not reached[Y]:
                stack.append([Y, step + 1])
                reached[x] = True
            if node == Y and not reached[X]:
                stack.append([X, step + 1])
                reached[x] = True
    return ret

if __name__ == '__main__':
    N, X, Y = map(int, input().split())
    for r in solve(N, X, Y)[1:]:
        print(r)

161 - Lunlun Number

Code 1. Tableau 3D

Quand j'ai vu le commentaire, j'ai entendu une voix disant "Ah". Il est trop tard pour le mettre en œuvre. time: O(N)

def solve(K):
    ref = [[['0'], ['1'], ['2'], ['3'], ['4'], ['5'], ['6'], ['7'], ['8'], ['9']]]
    if K < 10:
        return ref[0][K][0]

    rest = K - 9
    i = 0
    while True:
        i += 1
        ref.append([[] for _ in range(10)])
        for j in range(10):
            if j == 0:
                for r in ref[i-1][j] + ref[i-1][j+1]:
                    ref[i][j].append(str(j) + r)
            elif j == 9:
                for r in ref[i-1][j-1] + ref[i-1][j]:
                    ref[i][j].append(str(j) + r)
                    rest -= 1
                    if rest == 0:
                        return str(j) + r
            else:
                for r in ref[i-1][j-1] + ref[i-1][j] + ref[i-1][j+1]:
                    ref[i][j].append(str(j) + r)
                    rest -= 1
                    if rest == 0:
                        return str(j) + r

if __name__ == '__main__':
    K = int(input())
    print(solve(K))
Code 2. deque

C'était très rafraîchissant. time: O(N)

from collections import deque

def solve(K):
    stack = deque([1, 2, 3, 4, 5, 6, 7, 8, 9])
    for _ in range(K):
        num = stack.popleft()
        one_digit = num % 10
        if one_digit % 10 == 0:
            nexts = [num * 10, num * 10 + 1]
        elif one_digit % 10 == 9:
            nexts = [num * 10 + 8, num * 10 + 9]
        else:
            nexts = [num * 10 + one_digit - 1, num * 10 + one_digit, num * 10 + one_digit + 1]

        for n in nexts:
            stack.append(n)
    return num

if __name__ == '__main__':
    K = int(input())
    print(solve(K))

163 - Sum of Large Numbers

Code 1. Formule japonaise

time: O(N - K) space: O(1)

MOD = 10**9 + 7
class ModInt:
    def __init__(self, x):
        self.x = x % MOD

    def __str__(self):
        return str(self.x)

    __repr__ = __str__

    def __add__(self, other):
        return (
            ModInt(self.x + other.x) if isinstance(other, ModInt) else
            ModInt(self.x + other)
        )

    def __sub__(self, other):
        return (
            ModInt(self.x - other.x) if isinstance(other, ModInt) else
            ModInt(self.x - other)
        )

    def __mul__(self, other):
        return (
            ModInt(self.x * other.x) if isinstance(other, ModInt) else
            ModInt(self.x * other)
        )

    def __truediv__(self, other):
        return (
            ModInt(
                self.x * pow(other.x, MOD - 2, MOD)
            ) if isinstance(other, ModInt) else
            ModInt(self.x * pow(other, MOD - 2, MOD))
        )

    def __pow__(self, other):
        return (
            ModInt(pow(self.x, other.x, MOD)) if isinstance(other, ModInt) else
            ModInt(pow(self.x, other, MOD))
        )

    __radd__ = __add__

    def __rsub__(self, other):
        return (
            ModInt(other.x - self.x) if isinstance(other, ModInt) else
            ModInt(other - self.x)
        )

    __rmul__ = __mul__

    def __rtruediv__(self, other):
        return (
            ModInt(
                other.x * pow(self.x, MOD - 2, MOD)
            ) if isinstance(other, ModInt) else
            ModInt(other * pow(self.x, MOD - 2, MOD))
        )

    def __rpow__(self, other):
        return (
            ModInt(pow(other.x, self.x, MOD)) if isinstance(other, ModInt) else
            ModInt(pow(other, self.x, MOD))
        )
def sumOfArithmeticSequence(a, b):
    return (a + b) * (b - a + 1) // 2

def solve(N, K):
    ret = ModInt(0)
    for i in range(K, N + 2):
       ret += sumOfArithmeticSequence(N - i + 1, N) - sumOfArithmeticSequence(0, i - 1) + 1
    return ret


if __name__ == '__main__':
    N, K = map(int, input().split())
    print(solve(N, K))
Code 2. Somme cumulée

time: O(N) space: O(N)

MOD = 10**9 + 7
class ModInt:
    def __init__(self, x):
        self.x = x % MOD

    def __str__(self):
        return str(self.x)

    __repr__ = __str__

    def __add__(self, other):
        return (
            ModInt(self.x + other.x) if isinstance(other, ModInt) else
            ModInt(self.x + other)
        )

    def __sub__(self, other):
        return (
            ModInt(self.x - other.x) if isinstance(other, ModInt) else
            ModInt(self.x - other)
        )

    def __mul__(self, other):
        return (
            ModInt(self.x * other.x) if isinstance(other, ModInt) else
            ModInt(self.x * other)
        )

    def __truediv__(self, other):
        return (
            ModInt(
                self.x * pow(other.x, MOD - 2, MOD)
            ) if isinstance(other, ModInt) else
            ModInt(self.x * pow(other, MOD - 2, MOD))
        )

    def __pow__(self, other):
        return (
            ModInt(pow(self.x, other.x, MOD)) if isinstance(other, ModInt) else
            ModInt(pow(self.x, other, MOD))
        )

    __radd__ = __add__

    def __rsub__(self, other):
        return (
            ModInt(other.x - self.x) if isinstance(other, ModInt) else
            ModInt(other - self.x)
        )

    __rmul__ = __mul__

    def __rtruediv__(self, other):
        return (
            ModInt(
                other.x * pow(self.x, MOD - 2, MOD)
            ) if isinstance(other, ModInt) else
            ModInt(other * pow(self.x, MOD - 2, MOD))
        )

    def __rpow__(self, other):
        return (
            ModInt(pow(other.x, self.x, MOD)) if isinstance(other, ModInt) else
            ModInt(pow(other, self.x, MOD))
        )

def differenceOfMaximumSumAndMinimumSum(cumulative_sum, i):
    return (cumulative_sum[-1] - cumulative_sum[-i-1]) - cumulative_sum[i - 1]

def solve(N, K):
    ret = ModInt(1)
    cumulative_sum = [0] * (N + 1)
    tmp_sum = 0
    for i in range(1, N + 1):
        tmp_sum += i
        cumulative_sum[i] = tmp_sum

    for i in range(K, N + 1):
        ret += differenceOfMaximumSumAndMinimumSum(cumulative_sum, i) + 1
    return ret


if __name__ == '__main__':
    N, K = map(int, input().split())
    print(solve(N, K))

164 - Multiple of 2019

Code 1. DP

time: O(N) space: O(N)

MOD = 2019

def solve(S):
    N = len(S)
    dp = [0 for _ in range(MOD)]
    tmp = 0
    ret = 0
    for i in range(N - 1, -1, -1):
        m = (tmp + int(S[i]) * pow(10, N - i - 1, MOD)) % MOD
        ret += dp[m]
        dp[m] += 1
        tmp = m
    ret += dp[0]
    return ret

if __name__ == "__main__":
    S = input()
    print(solve(S))

167 - Teleporter

Code 1. Découverte du cycle de Floyd

time: O(N) space: O(N)

def solve(N, K, A):
    slow, fast = 1, 1
    is_first = True
    while is_first or slow != fast:
        is_first = False
        slow = A[slow - 1]
        fast = A[A[fast - 1] - 1]
    fast = 1
    steps_before_entering_cycle = []
    while slow != fast:
        steps_before_entering_cycle.append(fast)
        slow = A[slow - 1]
        fast = A[fast - 1]

    if K < len(steps_before_entering_cycle):
        return steps_before_entering_cycle[K]

    cycles = [slow]
    while A[slow - 1] != cycles[0]:
        cycles.append(A[slow - 1])
        slow = A[slow - 1]
    ret = cycles[(K - len(steps_before_entering_cycle)) % len(cycles)]
    return ret

if __name__ == '__main__':
    N, K = map(int, input().split())
    A = list(map(int, input().split()))
    print(solve(N, K, A))

168 - .. (Double Dots)

Code 1. BFS

time: O(N + M) space: O(N + M)

from collections import deque, defaultdict

def solve(N, M, paths):
    used = [False for _ in range(M)]
    stack = deque([1])
    ret = [[float("inf"), -1] for _ in range(N + 1)]
    ret[1] = [0, 1]

    while stack:
        cur = stack.popleft()
        nexts = paths[cur]
        for n in nexts:
            if used[n[0]]:
                continue
            used[n[0]] = True
            if ret[cur][0] + 1 < ret[n[1]][0]:
                ret[n[1]] = [ret[cur][0] + 1, cur]
                stack.append(n[1])

    print("Yes")
    for r in ret[2:]:
        print(r[1])

if __name__ == "__main__":
    N, M = map(int, input().split())
    paths = defaultdict(list)
    for idx in range(M):
        a, b = map(int, input().split())
        paths[a].append([idx, b])
        paths[b].append([idx, a])

    solve(N, M, paths)

Problème E

148 - Double Factorial

Code 1. Transformation d'expression

time: O(logN) space: O(1)

def solve(N):
   if N < 9 or N % 2:
       return 0
   ret = 0
   N //= 2
   while N:
       ret += N // 5
       N //= 5
   return ret

if __name__ == "__main__":
    N = int(input())
    print(solve(N))

153 - Crested Ibis vs Monster

Code 1. dp

time: O(NH) space: O(H)

def solve(h, n, A, B):
    dp = [float("inf")] * h + [0]
    for i in range(h, -1, -1):
        if dp[i] == float("inf"):
            continue
        for j in range(n):
            if 0 <= i - A[j]:
                dp[i - A[j]] = min(dp[i - A[j]], dp[i] + B[j])
            else:
                dp[0] = min(dp[0], dp[i] + B[j])
    return dp[0]

if __name__ == '__main__':
    h, n = map(int, input().split())
    a = [0] * n
    b = [0] * n

    for i in range(n):
        a[i], b[i] = map(int, input().split())

    print(solve(h, n, a, b))

160 - Red and Green Apples

Code 1. Trier + Itérer

time: max(O(X + Y), O(AlogA), O(BlogB), O(ClogC))

def solve(X, Y, p, q, r):
    red_apples = sorted(p, reverse=True)[: X]
    green_apples = sorted(q, reverse=True)[: Y]
    white_apples = sorted(r)

    ret = 0
    while red_apples and green_apples:
        if not white_apples or (white_apples[-1] < red_apples[-1] and white_apples[-1] < green_apples[-1]):
            return ret + sum(red_apples) + sum(green_apples)

        ret += white_apples.pop(-1)
        if red_apples[-1] < green_apples[-1]:
            red_apples.pop(-1)
        else:
            green_apples.pop(-1)


    if not red_apples:
        while green_apples and white_apples and (green_apples[-1] < white_apples[-1]):
            ret += white_apples.pop(-1)
            green_apples.pop(-1)
        return ret + sum(green_apples)

    else:
        while red_apples and white_apples and (red_apples[-1] < white_apples[-1]):
            ret += white_apples.pop(-1)
            red_apples.pop(-1)
        return ret + sum(red_apples)

if __name__ == '__main__':
    X, Y, A, B, C = map(int, input().split())
    p = list(map(int, input().split()))
    q = list(map(int, input().split()))
    r = list(map(int, input().split()))
    print(solve(X, Y, p, q, r))
Code 2. Trier

C'est plus concis. time: max(O(AlogA), O(BlogB), O((X + Y + C)log(X + Y + C)))

def solve(X, Y, p, q, r):
    red_apples = sorted(p, reverse=True)[: X]
    green_apples = sorted(q, reverse=True)[: Y]
    return sum(sorted(red_apples + green_apples + r, reverse=True)[: X + Y])

if __name__ == '__main__':
    X, Y, A, B, C = map(int, input().split())
    p = list(map(int, input().split()))
    q = list(map(int, input().split()))
    r = list(map(int, input().split()))
    print(solve(X, Y, p, q, r))

166 - This Message Will Self-Destruct in 5s

Code 1. Italate + Dictionnaire

time: O(N) sapce: O(N)

def solve(N, A):
    ret = 0
    ref = {}
    for idx, a in enumerate(A):
        ith_val = idx - a
        ret += ref[ith_val] if ith_val in ref else 0
        ref[idx + a] = ref.get(idx + a, 0) + 1
    return ret

if __name__ == '__main__':
    N = int(input())
    A = list(map(int, input().split()))
    print(solve(N, A))

167 - Colorful Blocks

Code 1. Mettez en italique le nombre de couleurs différentes

time: O(N) space: O(N)

MOD = 998244353
class ModInt:
    def __init__(self, x):
        self.x = x % MOD

    def __str__(self):
        return str(self.x)

    __repr__ = __str__

    def __add__(self, other):
        return (
            ModInt(self.x + other.x) if isinstance(other, ModInt) else
            ModInt(self.x + other)
        )

    def __sub__(self, other):
        return (
            ModInt(self.x - other.x) if isinstance(other, ModInt) else
            ModInt(self.x - other)
        )

    def __mul__(self, other):
        return (
            ModInt(self.x * other.x) if isinstance(other, ModInt) else
            ModInt(self.x * other)
        )

    def __truediv__(self, other):
        return (
            ModInt(
                self.x * pow(other.x, MOD - 2, MOD)
            ) if isinstance(other, ModInt) else
            ModInt(self.x * pow(other, MOD - 2, MOD))
        )

    def __pow__(self, other):
        return (
            ModInt(pow(self.x, other.x, MOD)) if isinstance(other, ModInt) else
            ModInt(pow(self.x, other, MOD))
        )

    __radd__ = __add__

    def __rsub__(self, other):
        return (
            ModInt(other.x - self.x) if isinstance(other, ModInt) else
            ModInt(other - self.x)
        )

    __rmul__ = __mul__

    def __rtruediv__(self, other):
        return (
            ModInt(
                other.x * pow(self.x, MOD - 2, MOD)
            ) if isinstance(other, ModInt) else
            ModInt(other * pow(self.x, MOD - 2, MOD))
        )

    def __rpow__(self, other):
        return (
            ModInt(pow(other.x, self.x, MOD)) if isinstance(other, ModInt) else
            ModInt(pow(other, self.x, MOD))
        )

def cmb(fact, n ,r):
    return ModInt(1) * fact[n] / fact[n - r] / fact[r]

def solve(N, M, K):
    fact = [ModInt(1)]
    for i in range(1, N + 1):
        fact.append(fact[i -1] * i)

    ret = ModInt(0)
    for x in range(K + 1):
        ret += cmb(fact, N - 1, x) * M * pow(M - 1, N - x - 1, MOD)
    return ret

if __name__ == '__main__':
    N, M, K = map(int, input().split())
    print(solve(N, M, K))

ARC

Problème B

26-Numéro complet

Code 1. Énumérer les réductions

time: O(\sqrt{N}) space: O(\sqrt{N})

def getDivs(N):
    divs = []
    for i in range(1, int(N**0.5) + 1):
        if not N % i:
            divs.append(i)
            if i != N // i:
                divs.append(N // i)
    return divs

def solve(N):
    divs = getDivs(N)
    div_sum = sum(divs)
    if div_sum == 2 * N:
        return "Perfect"
    return "Abundant" if 2 * N < div_sum else "Deficient"

if __name__ == "__main__":
    N = int(input())
    print(solve(N))

AGC

Un problème

43 - Range Flip Find Route

Code 1. DP

C'était plus difficile à considérer que la mise en œuvre. time: O(HW) space: O(HW)

def solve(H, W, s):
    scores = [[float('inf') for _ in range(W)] for _ in range(H)]
    scores[0][0] = 0 if s[0][0] == '.' else 1

    for i in range(H):
        for j in range(W):
            for x, y in [(i - 1, j), (i, j - 1)]:
                if x < 0 or y < 0:
                    continue
                if s[i][j] == '.' or s[x][y] == '#':
                    scores[i][j] = min(scores[i][j], scores[x][y])
                else:
                    scores[i][j] = min(scores[i][j], scores[x][y] + 1)

    return scores[H - 1][W - 1]

if __name__ == '__main__':
    H, W = map(int, input().split())
    s = []
    for _ in range(H):
        s.append(list(input()))

    print(solve(H, W, s))

Problème B

3 - Simplified mahjong

Code 1. Méthode gourmande

time: O(N) space: O(1)

def solve(N, A):
    ret = 0
    for idx in range(N):
        ret += A[idx] // 2
        A[idx] %= 2
        if A[idx] and idx + 1 < N and A[idx + 1]:
            ret += 1
            A[idx + 1] -= 1
    return ret

if __name__ == "__main__":
    N = int(input())
    A = [0 for _ in range(N)]
    for idx in range(N):
        A[idx] = int(input())
    print(solve(N, A))

Autre

Problème B

Concours de programmation Keyence 2020 - Robot Arms

Code 1. Méthode gourmande

time: O(N) sapce: O(1)

def solve(N, data):
   ret = N
   data.sort(key=lambda x: x[0] - x[1])
   most_right = float("-inf")
   for idx, d in enumerate(data):
       l, r = d[0] - d[1], d[0] + d[1]
       if not idx or most_right <= l:
           most_right = r
           continue
       ret -= 1
       most_right = min(r, most_right)
   return ret

if __name__ == "__main__":
    N = int(input())
    data = []

    for _ in range(N):
        data.append(list(map(int, input().split())))
    print(solve(N, data))

Problème C

SoundHound Inc. Programming Contest 2018 - Ordinary Beauty

Pour moi, une première connaissance complète était nécessaire. Je pourrais penser à une solution pour $ O (m) $ avec $ d = 1 $, mais je ne pouvais pas la généraliser et l'accélérer.

Code 1. Linéarité de la valeur attendue

Explication de la "linéarité des valeurs attendues" et résumé des problèmes d'utilisation time: O(1) sapce: O(1)

def solve(n, m, d):
    return (m - 1) * 2 * (n - d) / n**2 if d else (m - 1) / n

if __name__ == "__main__":
    n, m, d = map(int, input().split())
    print(solve(n, m, d))

Problème D

Concours de programmation Sumitomo Mitsui Trust Bank 2019 - PIN porte-bonheur

Code 1. Italate + Dictionnaire

time: O(100\ N) sapce: O(100\ N) 100\ N \leq 3\times 10^6

import copy
def solve(N, S):
    single_ref = [False for _ in range(10)]
    double_ref = [{} for _ in range(N)]

    for i in range(N - 1, -1, -1):
        if i == N - 1:
            single_ref[int(S[i])] = True
            continue
        double_ref[i] = copy.deepcopy(double_ref[i + 1])
        for j in range(10):
            if not single_ref[j]:
                continue
            double_ref[i][S[i] + str(j)] = 1
        single_ref[int(S[i])] = True

    ret = {}
    for i in range(N - 1):
        for k in double_ref[i + 1].keys():
            ret[S[i] + k] = 1
    return len(ret.keys())

if __name__ == "__main__":
    N = int(input())
    S = input()
    print(solve(N, S))

Concours de programmation Panasonic 2020 - Équivalence de chaîne

Code 1. DFS
def dfs(s, mx):
    if len(s) == N:
        print(s)
        return
    for c in range(ord("a"), mx + 1):
        dfs(s + chr(c), mx + 1 if c == mx else mx)

if __name__ == "__main__":
    N = int(input())
    dfs("", ord("a"))

Problème E - Chapeaux colorés 2

Concours de programmation Sumitomo Mitsui Trust Bank 2019

Code 1. Italate

time: O(N) space: O(N)

MOD = 10**9 + 7
class ModInt:
    def __init__(self, x):
        self.x = x % MOD

    def __str__(self):
        return str(self.x)

    __repr__ = __str__

    def __add__(self, other):
        return (
            ModInt(self.x + other.x) if isinstance(other, ModInt) else
            ModInt(self.x + other)
        )

    def __sub__(self, other):
        return (
            ModInt(self.x - other.x) if isinstance(other, ModInt) else
            ModInt(self.x - other)
        )

    def __mul__(self, other):
        return (
            ModInt(self.x * other.x) if isinstance(other, ModInt) else
            ModInt(self.x * other)
        )

    def __truediv__(self, other):
        return (
            ModInt(
                self.x * pow(other.x, MOD - 2, MOD)
            ) if isinstance(other, ModInt) else
            ModInt(self.x * pow(other, MOD - 2, MOD))
        )

    def __pow__(self, other):
        return (
            ModInt(pow(self.x, other.x, MOD)) if isinstance(other, ModInt) else
            ModInt(pow(self.x, other, MOD))
        )

    __radd__ = __add__

    def __rsub__(self, other):
        return (
            ModInt(other.x - self.x) if isinstance(other, ModInt) else
            ModInt(other - self.x)
        )

    __rmul__ = __mul__

    def __rtruediv__(self, other):
        return (
            ModInt(
                other.x * pow(self.x, MOD - 2, MOD)
            ) if isinstance(other, ModInt) else
            ModInt(other * pow(self.x, MOD - 2, MOD))
        )

    def __rpow__(self, other):
        return (
            ModInt(pow(other.x, self.x, MOD)) if isinstance(other, ModInt) else
            ModInt(pow(other, self.x, MOD))
        )

def solve(N, A):
    ret = ModInt(1)
    count = [3 if not i else 0 for i in range(N + 1)]

    for a in A:
        ret *= count[a]
        if not ret:
            break
        count[a] -= 1
        count[a + 1] += 1

    return ret

if __name__ == "__main__":
    N = int(input())
    A = list(map(int, input().split()))
    print(solve(N, A))

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