AtCoder Beginner Contest 297 【E-F】题解

E - Kth Takoyaki Set

多重指针，具体可参考leetcode简化版题目丑数

import math
import sys
from bisect import bisect_left, bisect_right
from collections import Counter, defaultdict, deque
from itertools import permutationsinput = lambda: sys.stdin.readline().rstrip("\r\n")def I():return input()def ii():return int(input())def li():return list(input().split())def mi():return map(int, input().split())def lii():return list(map(int, input().split()))# 8 3
# 5 3
# 2 3
# 2 1
# 1 1n,k = mi()
a = lii()
dp = [0] * (k + 1)
pois = [0] * n
i = 1
while i <= k:mi = float('inf')for j in range(n):mi = min(mi,dp[pois[j]] + a[j])for j in range(n):if mi == dp[pois[j]] + a[j]:dp[i] = mipois[j] += 1i += 1
print(dp[k])

F - Minimum Bounding Box 2

枚举矩形的高h和宽w，通过容斥枚举不能够组成该矩形的充要条件（即有一条边没有点），一共四种情况，最上面少一条边，最下面少一条边，最左边少一天边，最右边少一条边。
然后减去重复的，此时是四容斥情况
具体容斥公式可见OI Wiki进行推导

import math
import sys
from bisect import bisect_left, bisect_right
from collections import Counter, defaultdict, deque
from itertools import permutationsinput = lambda: sys.stdin.readline().rstrip("\r\n")def I():return input()def ii():return int(input())def li():return list(input().split())def mi():return map(int, input().split())def lii():return list(map(int, input().split()))def qmi(x, y, mod):res = 1while y:if y & 1: res = res * x % modx = x * x % mody >>= 1return res % modclass Factorial:def __init__(self, N, mod) -> None:N += 1self.mod = modself.f = [1 for _ in range(N)]self.g = [1 for _ in range(N)]for i in range(1, N):self.f[i] = self.f[i - 1] * i % self.modself.g[-1] = pow(self.f[-1], mod - 2, mod)for i in range(N - 2, -1, -1):self.g[i] = self.g[i + 1] * (i + 1) % self.moddef fac(self, n):return self.f[n]def fac_inv(self, n):return self.g[n]def comb(self, n, m):if n < m or m < 0:return 0return self.f[n] * self.g[m] % self.mod * self.g[n - m] % self.moddef perm(self, n, m):if n < m or m < 0:return 0return self.f[n] * self.g[n - m] % self.moddef catalan(self, n):return (self.comb(2 * n, n) - self.comb(2 * n, n - 1)) % self.moddef inv(self, n):return self.f[n - 1] * self.g[n] % modmod = 998244353
fact = Factorial(10 ** 6 + 3, mod)def comb(x, y, k):if x <= 0 or y <= 0 or k < 0 or x * y < k: return 0return fact.comb(x * y, k)h, w, k = lii()ans = 0
for n in range(1, h + 1):for m in range(1, w + 1):cnt = (h - n + 1) * (w - m + 1)q = comb(n, m, k) - comb(n, (m - 1), k) * 2 - comb(n - 1, m, k) * 2 \+ comb((n - 2), m, k) + comb(n, (m - 2), k) + 4 * comb((n - 1), (m - 1), k) \- comb((n - 2), (m - 1), k) * 2 - comb((n - 1), (m - 2), k) * 2 + comb((n - 2), (m - 2), k)ans += q * cnt * n * mans %= mod
print(ans * qmi(fact.comb(h * w, k), mod - 2, mod) % mod)

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AtCoder Beginner Contest 297 【E-F】题解

AtCoder Beginner Contest 297 【E-F】题解

E - Kth Takoyaki Set

F - Minimum Bounding Box 2

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