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Neko Finds Grapes
#include <bits/stdc++.h>
using namespace std;
int n, m, a[2], b[2];
int main()
{
scanf("%d%d", &n, &m);
for (int i = 0, t; i < n; ++i)
scanf("%d", &t), ++a[t & 1];
for (int i = 0, t; i < m; ++i)
scanf("%d", &t), ++b[t & 1];
printf("%d", min(a[0], b[1]) + min(a[1], b[0]));
}
Neko Performs Cat Furrier Transform
每次把最高位异或掉即可。
#include <bits/stdc++.h>
using namespace std;
int x, c, b[30], ans[40], *p;
int main()
{
for (int i = 0; i < 30; ++i)
b[i] = (1 << i) - 1;
for (scanf("%d", &x); p = lower_bound(b, b + 30, x), *p != x; ++c)
{
if (c & 1)
++x;
else
x ^= *p, ans[c] = p - b;
}
printf("%d\n", c);
for (int i = 0; i < c; i += 2)
printf("%d ", ans[i]);
}
Neko does Maths
题意：找到最小的非负整数$k$，使得$(k+a)/\gcd(k+a,k+b)\times(k+b)$最小。
解：不妨$a\ge b$，考虑$\gcd(a,b)=\gcd(a,a-b)$，用$c=a-b$代入原式，即要最小化$(k+a)/\gcd(k+a,k+a-c)\times(k+a-c)=(k+a)/\gcd(k+a,c)\times(k+a-c)$。由于$\gcd(k+a,c)$一定是$c$的因子，枚举$c$的所有因子$j$，要求$j$也是$k+a$的因子，满足的$k$在$[0,j)$内存在且唯一，此时回代判断即可。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a, b;
int main()
{
scanf("%lld%lld", &a, &b);
if (a < b)
swap(a, b);
pair<ll, ll> ans(a / __gcd(a, b) * b, 0);
for (ll i = 1, c = a - b; i * i <= c; ++i)
if (c % i == 0)
for (auto j : {i, c / i})
{
ll k = ((-a) % j + j) % j;
ans = min(ans, {(k + a) / j * (k + a - c), k});
}
printf("%lld", ans.second);
}
Neko and Aki’s Prank
题意：有一个 trie 是从一个括号序列建立的，求这个 trie 的最大匹配是多少。
解：现场已经看出这个题是树的 DP，我还是太菜了啊。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
struct Mod
{
const ll M, SM;
Mod(ll M) : M(M), SM(sqrt(M) + 0.5) {}
ll qadd(ll a, ll b) const { return a += b, a < M ? a : a - M; } //假如a和b都已经在同余系内，就不必取模了，取模运算耗时很高
ll add(ll a, ll b) const { return a = (a + b) % M, a < 0 ? a + M : a; } //考虑a和b不在同余系内甚至为负数的情况
ll mul(ll a, ll b) const { return add(a * b, 0); }
/*
ll mul(ll x, ll y) const //无循环快速计算同余乘法，根据a*b是否爆ll替换a*b%m
{
ll a = x / SM, b = x % SM,
c = y / SM, d = y % SM,
e = SM * SM - M, //可能为负
f = ((a * d + b * c) % M + a * c / SM * e) % M;
return add(((a * c % SM + f / SM) * e % M + f % SM * SM) % M, b * d);
}
*/
ll pow(ll a, ll b) const
{
ll r = 1;
for (a = add(a, 0LL); b; b >>= 1, a = mul(a, a))
if (b & 1)
r = mul(r, a);
return r;
}
ll inv(ll a) const { return pow(a, M - 2); } //return pow(a, phi(M) - 1);
/*
ll inv(ll a) const //模m下a的乘法逆元，不存在返回-1（m为素数时a不为0必有逆元）
{
ll x, y, d = gcd(a, m, x, y);
return d == 1 ? (x + m) % m : -1;
}
*/
ll log(ll a, ll b) const
{
unordered_map<ll, ll> x;
for (ll i = 0, e = 1; i <= SM; ++i, e = mul(e, a))
if (!x.count(e))
x[e] = i;
for (ll i = 0, v = inv(pow(a, SM)); i <= SM; ++i, b = mul(b, v))
if (x.count(b))
return i * SM + x[b];
return -1;
}
/*
vector<ll> sol(ll a, ll b) const //解同余方程，返回ax=b(mod M)循环节内所有解
{
vector<ll> ans;
ll x, y, d = gcd(a, M, x, y);
if (b % d)
return ans;
ans.push_back((b / d) % (M / d) * (x = (x % M + M) % M));
for (ll i = 1; i < d; ++i)
ans.push_back((ans.back() + M / d) % M);
return ans;
}
*/
} M(1e9 + 7);
const ll N = 2047;
ll n, f[N][N][2];
int main()
{
memset(f, -0x3f, sizeof(f));
f[0][0][0] = 0;
for (int i = 1; i + 1 < N; ++i)
{
f[i][0][0] = max(f[i - 1][1][0], f[i - 1][1][1]);
f[i][0][1] = M.qadd(f[i - 1][1][0], 1);
for (int j = 1; j <= i; ++j)
{
f[i][j][0] = M.qadd(
max(0LL, max(f[i - 1][j - 1][0], f[i - 1][j - 1][1])),
max(0LL, max(f[i - 1][j + 1][0], f[i - 1][j + 1][1])));
f[i][j][1] = max(
f[i - 1][j - 1][0] + 1 + max(0LL, max(f[i - 1][j + 1][1], f[i - 1][j + 1][0])),
f[i - 1][j + 1][0] + 1 + max(0LL, max(f[i - 1][j - 1][0], f[i - 1][j - 1][1]))) %
M.M;
}
}
scanf("%lld", &n);
printf("%lld", max(f[2 * n][0][0], f[2 * n][0][1]));
}
Neko and Flashback
欧拉路。详见官方题解。
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 9;
int n, b[N], c[N];
struct Ranker : vector<int>
{
void init() { sort(begin(), end()), resize(unique(begin(), end()) - begin()); }
int ask(int x) { return lower_bound(begin(), end(), x) - begin(); }
} rk;
struct Graph
{
struct Vertex
{
vector<int> i;
};
typedef pair<int, int> Edge;
vector<Vertex> v;
vector<Edge> e;
Graph(int n) : v(n) {}
void add(const Edge &ed)
{
v[ed.second].i.push_back(e.size());
e.push_back(ed);
}
};
struct Fleury : Graph
{
vector<int> vis, cur, p;
Fleury(int n) : Graph(n) {}
void dfs(int u)
{
for (int &i = cur[u], k; i < v[u].i.size(); ++i)
if (k = v[u].i[i], !vis[k] && !vis[k ^ 1])
vis[k] = 1, dfs(e[k].first), p.push_back(k);
}
void ask()
{
int cnt = 0;
for (int i = 0; i < v.size(); ++i)
if (v[i].i.size() % 2)
++cnt;
if (cnt > 2)
return;
vis.assign(e.size(), 0), cur.assign(v.size(), 0), p.clear();
for (int i = 0; i < v.size(); ++i)
if (v[i].i.size() % 2)
return dfs(i);
dfs(0);
}
};
int main()
{
scanf("%d", &n);
for (int i = 0; i < n - 1; ++i)
scanf("%d", &b[i]), rk.push_back(b[i]);
for (int i = 0; i < n - 1; ++i)
scanf("%d", &c[i]), rk.push_back(c[i]);
rk.init();
Fleury g(rk.size());
for (int i = 0; i < n - 1; ++i)
{
if (b[i] > c[i])
return printf("-1"), 0;
g.add({b[i] = rk.ask(b[i]), c[i] = rk.ask(c[i])}), g.add({c[i], b[i]});
}
g.ask();
if (g.p.size() < n - 1)
return printf("-1"), 0;
for (int i = 0; i < n - 1; ++i)
printf("%d ", rk[g.e[g.p[i]].first]);
printf("%d ", rk[g.e[g.p[n - 2]].second]);
}
Neko Rules the Catniverse (Small Version)
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$mask$。
这里可以滚动数组。
#include <bits/stdc++.h>
using namespace std;
void add(int &a, long long b, int M = 1e9 + 7) { a = (a + b) % M; }
int n, k, m;
int main()
{
scanf("%d%d%d", &n, &k, &m);
vector<vector<int>> f(k + 1, vector<int>(1 << m, 0));
f[0][0] = 1;
for (int i = 0; i < n; ++i)
{
vector<vector<int>> g(k + 1, vector<int>(1 << m, 0));
for (int j = 0; j < f.size(); ++j)
for (int mask = 0; mask < f[j].size(); ++mask)
{
int newMask = (mask << 1) % f[j].size();
add(g[j][newMask], f[j][mask]);
if (j < k)
add(g[j + 1][newMask | 1], (1LL + __builtin_popcount(mask)) * f[j][mask]);
}
f.swap(g);
}
for (int mask = m = 0; mask < f[k].size(); ++mask)
add(m, f[k][mask]);
printf("%d", m);
}
Neko Rules the Catniverse (Large Version)
注意到状态只在相邻的两位转移，这里可以用矩阵加速 DP。
下面这段代码跑了 5070ms，官方还有一个跑了 156ms 的解法，暂时没看懂…也太强了吧。
#include <bits/stdc++.h>
using namespace std;
int n, k, m;
void add(int &a, long long b, int M = 1e9 + 7) { a = (a + b) % M; }
int toId(int j, int mask) { return j << m | mask; }
struct Matrix
{
vector<vector<int>> a;
int n;
Matrix(int n) : n(n), a(n, vector<int>(n)) {}
friend Matrix operator*(const Matrix &a, const Matrix &b)
{
Matrix c(a.n);
for (int i = 0; i < a.n; ++i)
for (int j = 0; j < a.n; ++j)
for (int k = 0; k < a.n; ++k)
add(c.a[i][j], 1LL * a.a[i][k] * b.a[k][j]);
return c;
}
friend Matrix pow(Matrix a, int b)
{
Matrix r(a.n);
for (int i = 0; i < a.n; ++i)
r.a[i][i] = 1;
for (; b; b >>= 1, a = a * a)
if (b & 1)
r = a * r;
return r;
}
};
int main()
{
scanf("%d%d%d", &n, &k, &m);
Matrix dp(toId(k + 1, 0)), f(dp);
for (int j = 0; j <= k; ++j)
for (int mask = 0; mask < (1 << m); ++mask)
{
int newMask = (mask << 1) % (1 << m);
f.a[toId(j, newMask)][toId(j, mask)] = 1;
if (j < k)
f.a[toId(j + 1, newMask | 1)][toId(j, mask)] = 1 + __builtin_popcount(mask);
}
dp.a[0][0] = 1, dp = pow(f, n) * dp;
int ans = 0;
for (int mask = 0; mask < (1 << m); ++mask)
add(ans, dp.a[toId(k, mask)][0]);
printf("%d", ans);
}