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wu-kan
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v2.1.3 页面样式微调,将masthead调矮,将标题字号改小
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README.md

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@@ -76,8 +76,8 @@ Live2D的后端暂时使用了[fghrsh](https://www.fghrsh.net/post/170.html)提
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> <p style="text-indent:2em">章北海感到父亲的灵魂从冥冥中降落到飞船上,与他融为一体,他按动了操作界面上那个最后的按钮,心中默念出那个他用尽一生的努力所追求的指令:</p><p style="text-indent:2em">“‘自然选择’,前进四!”</p>
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## 历程
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### 2019-04-12 v2.1.3
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页面样式微调,将masthead调矮
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### 2019-04-29 v2.1.3
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页面样式微调,将masthead调矮,将标题字号改小
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### 2019-03-20 v2.1.2
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修复sidebar展开时回到顶部的问题
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### 2019-03-18 v2.1.1

_config.yml

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@@ -95,12 +95,6 @@ head: # 删改前确定知道自己在做什么
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value: stylesheet
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- key: href
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value: /public/css/lanyon.css
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- tag: link
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var:
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- key: rel
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value: stylesheet
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- key: href
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value: //fonts.loli.net/css?family=PT+Serif:400,400italic,700%7CPT+Sans:400
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- tag: link
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var:
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- key: rel

_layouts/page.html

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<!-- Wrap is the content to shift when toggling the sidebar. We wrap the
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content to avoid any CSS collisions with our real content. -->
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<div class="wrap">
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<h3 class="masthead masthead-title">
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<h1 class="masthead masthead-title">
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<div class="container">
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{{ page.title }}
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<a href="{{ site.baseurl }}/">
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<small>{{ site.title }}</small>
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</a>
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</div>
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</h3>
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</h1>
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<div class="container content">
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{{ content }}
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</div>

_posts/2019-04-25-Codeforces Round #554 (Div. 2).md

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---
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title: Codeforces Round \#554 (Div. 2)
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title: Codeforces Round #554 (Div. 2)
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categories:
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- ACM
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- 题解
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printf("%d ", rk[g.e[g.p[n - 2]].second]);
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}
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```
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# [Neko Rules the Catniverse (Small Version)](https://vjudge.net/problem/CodeForces-1152F1)
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状压DP,用$f[i][j][mask]$来表示正在考虑第i个,已经考虑了第$j$个,后$m$个的选择状态是二进制位集$mask$。
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这里可以滚动数组。
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```cpp
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#include <bits/stdc++.h>
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using namespace std;
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void add(int &a, long long b, int M = 1e9 + 7) { a = (a + b) % M; }
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int n, k, m;
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int main()
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{
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scanf("%d%d%d", &n, &k, &m);
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vector<vector<int>> f(k + 1, vector<int>(1 << m, 0));
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f[0][0] = 1;
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for (int i = 0; i < n; ++i)
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{
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vector<vector<int>> g(k + 1, vector<int>(1 << m, 0));
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for (int j = 0; j < f.size(); ++j)
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for (int mask = 0; mask < f[j].size(); ++mask)
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{
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int newMask = (mask << 1) % f[j].size();
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add(g[j][newMask], f[j][mask]);
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if (j < k)
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add(g[j + 1][newMask | 1], (1LL + __builtin_popcount(mask)) * f[j][mask]);
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}
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f.swap(g);
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}
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for (int mask = m = 0; mask < f[k].size(); ++mask)
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add(m, f[k][mask]);
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printf("%d", m);
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}
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```
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# [Neko Rules the Catniverse (Large Version)](https://vjudge.net/problem/CodeForces-1152F2)
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注意到状态只在相邻的两位转移,这里可以用矩阵加速DP。
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下面这段代码跑了5070ms,官方还有一个[跑了156ms的解法](https://codeforces.com/contest/1152/submission/53260183),暂时没看懂…也太强了吧。
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```cpp
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#include <bits/stdc++.h>
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using namespace std;
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int n, k, m;
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void add(int &a, long long b, int M = 1e9 + 7) { a = (a + b) % M; }
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int toId(int j, int mask) { return j << m | mask; }
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struct Matrix
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{
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vector<vector<int>> a;
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int n;
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Matrix(int n) : n(n), a(n, vector<int>(n)) {}
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friend Matrix operator*(const Matrix &a, const Matrix &b)
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{
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Matrix c(a.n);
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for (int i = 0; i < a.n; ++i)
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for (int j = 0; j < a.n; ++j)
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for (int k = 0; k < a.n; ++k)
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add(c.a[i][j], 1LL * a.a[i][k] * b.a[k][j]);
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return c;
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}
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friend Matrix pow(Matrix a, int b)
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{
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Matrix r(a.n);
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for (int i = 0; i < a.n; ++i)
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r.a[i][i] = 1;
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for (; b; b >>= 1, a = a * a)
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if (b & 1)
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r = a * r;
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return r;
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}
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};
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int main()
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{
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scanf("%d%d%d", &n, &k, &m);
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Matrix dp(toId(k + 1, 0)), f(dp);
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for (int j = 0; j <= k; ++j)
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for (int mask = 0; mask < (1 << m); ++mask)
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{
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int newMask = (mask << 1) % (1 << m);
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f.a[toId(j, newMask)][toId(j, mask)] = 1;
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if (j < k)
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f.a[toId(j + 1, newMask | 1)][toId(j, mask)] = 1 + __builtin_popcount(mask);
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}
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dp.a[0][0] = 1, dp = pow(f, n) * dp;
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int ans = 0;
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for (int mask = 0; mask < (1 << m); ++mask)
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add(ans, dp.a[toId(k, mask)][0]);
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printf("%d", ans);
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}
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```

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