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CVBB ACM Team 2020-2021:teams:alchemist:mountvoom https://wiki.cvbbacm.com/ 2025-07-05T13:59:01+0800 CVBB ACM Team https://wiki.cvbbacm.com/ https://wiki.cvbbacm.com/lib/exe/fetch.php?media=favicon.ico text/html 2020-05-07T17:35:00+0800 Anonymous (anonymous@undisclosed.example.com) 2020-2021:teams:alchemist:mountvoom:graphtheory https://wiki.cvbbacm.com/doku.php?id=2020-2021:teams:alchemist:mountvoom:graphtheory&rev=1588844100&do=diff 霍尔定理 text/html 2020-05-12T10:29:02+0800 Anonymous (anonymous@undisclosed.example.com) 2020-2021:teams:alchemist:mountvoom:halltheorem https://wiki.cvbbacm.com/doku.php?id=2020-2021:teams:alchemist:mountvoom:halltheorem&rev=1589250542&do=diff 霍尔定理 设二分图的两部分为$X$、$Y$,且$|X|\leq|Y|$。 则定理描述为:二分图存在完美匹配,等价于对于$X$的任意子集$X^{'}$,与它们中任意点相连的$Y$的结点个数$\ge |X^{'}|$。 gym102268D Dates 题意: 给你一张二分图,左边有$t$个位置,右边右$n$$i$$[l_i, r_i]$$1 \leq n, t \leq 3 \times 10^5, l_i \leq l_{i + 1}, r_i \leq r_{i + 1}$$l_i \leq l_{i + 1}, r_i \leq r_{i + 1}$$\forall 1\le i< j\le n,[i,j]\text{中被选择的右侧点个数}\le [l_i,r_j]\text{中左侧点数量}$$pre[i]$$a[i]$$p[i]$$[1, i]$$\forall 1\le i< j\le n, p[j]-p[i-1]\le pre[r_j]-pre[l_i-1]$$\forall 1\le i< j\le n, pre[l_i-1]-p[i-1]\le pre[r_j]… text/html 2020-05-13T19:08:16+0800 Anonymous (anonymous@undisclosed.example.com) 2020-2021:teams:alchemist:mountvoom:myself https://wiki.cvbbacm.com/doku.php?id=2020-2021:teams:alchemist:mountvoom:myself&rev=1589368096&do=diff 个人训练 同济大学程序设计预选赛 text/html 2020-05-13T20:25:06+0800 Anonymous (anonymous@undisclosed.example.com) 2020-2021:teams:alchemist:mountvoom:training1 https://wiki.cvbbacm.com/doku.php?id=2020-2021:teams:alchemist:mountvoom:training1&rev=1589372706&do=diff 简况 AC 5题,属实菜逼。 比赛链接 题解 A. 张老师和菜哭武的游戏 题意: 有$1 \sim n$共$n$个数,最开始拿走$a, b, a \ne b$,当数$j$能被拿走时,当且仅当$\exists x, y$满足$x, y$已经被拿走且$x + y = j$或$x - y = j$,判断能拿走的数的个数的奇偶性。 题解: 可以看出,能被拿走的数一定能用$x * a + y * b$$gcd(a, b)$$n / gcd(a, b)$$n, n \leq 10^3$$x$$i$$p_i$$t_i$$dp[i][j][0/1]$$[i, j]$$k = 0$$i$$j$$dp[i + 1][j][0/1], dp[i][j - 1][0/1]$$w \times h$$k, k \leq 5$$2N, N \leq 10^7$$N$$\frac{(2n)!}{n! \times 2^n}$$f(n) = \sum f(i) * f(n - i - 1) = \frac{C(2n, n)}{n + 1}$$[l, r]$$1 \sim n$$0, t, v$$t$$…