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Adrien and Austin
特判$n=0$和$k=1$的两种情况。对于其他情况，先手方可以通过取1或2将原始序列断成左右等长的两部分，然后「模仿」对手的操作即必胜。
#include <bits/stdc++.h>
using namespace std;
int main()
{
int n, k;
scanf("%d%d", &n, &k);
if (n == 0)
return printf("Austin"), 0;
if (k == 1)
return printf(n & 1 ? "Adrien" : "Austin"), 0;
printf("Adrien");
}
Country Meow
三维最小球覆盖模板题，这里拉了一个模拟退火的模板。
#include <bits/stdc++.h>
using namespace std;
typedef double lf;
const lf EPS = 1e-9, INF = 1e9;
int sgn(lf d) { return (d > EPS) - (d < -EPS); }
struct Coord3
{
lf X, Y, Z;
Coord3 &operator-=(const Coord3 &b) { return X -= b.X, Y -= b.Y, Z -= b.Z, *this; }
friend Coord3 operator-(Coord3 a, const Coord3 &b) { return a -= b; }
Coord3 &operator*=(lf d) { return X *= d, Y *= d, Z *= d, *this; }
friend Coord3 operator*(Coord3 a, lf d) { return a *= d; }
friend Coord3 operator*(lf d, Coord3 a) { return a *= d; }
friend lf Dot(const Coord3 &A, const Coord3 &B) { return A.X * B.X + A.Y * B.Y + A.Z * B.Z; }
friend lf norm(const Coord3 &A) { return Dot(A, A); }
friend lf abs(const Coord3 &A) { return sqrt(norm(A)); }
friend lf minBall(const vector<Coord3> &data, const lf eps = EPS * 1e-3) //模拟退火求最小球覆盖，EPS玄学调整
{
lf step = 1, ans = INF;
for (Coord3 z{0, 0, 0}; step > eps; step *= 0.99)
{
int s = 0;
for (int i = 0; i < data.size(); ++i)
if (abs(z - data[s]) < abs(z - data[i]))
s = i;
ans = min(ans, abs(z - data[s]));
z -= (z - data[s]) * step;
}
return ans;
}
};
int main()
{
int n;
scanf("%d", &n);
vector<Coord3> p(n);
for (int i = 0; i < n; ++i)
scanf("%lf%lf%lf", &p[i].X, &p[i].Y, &p[i].Z);
printf("%.9f\n", minBall(p));
}
Pyramid
打表找规律后秒掉。要注意的是这里除法逆元要预处理，否则超时。
#include <bits/stdc++.h>
using namespace std;
typedef int ll;
struct Mod
{
const ll M;
Mod(const ll M) : M(M) {}
ll mul(ll a, long long b) const { return a * b % M; }
ll pow(ll a, ll b) const
{
ll r = 1;
for (a %= M; 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); }
} M(1e9 + 7);
ll t, n, inv = M.inv(24);
int main()
{
for (scanf("%d", &t); t--;)
{
scanf("%d", &n);
printf("%d\n", M.mul(n - 3, M.mul(n - 4, M.mul(n - 5, M.mul(n - 6, inv)))));
}
}
Magic Potion
想到网络流之后就很好建图了。
#include <bits/stdc++.h>
using namespace std;
typedef int ll;
const ll INF = 1e9;
struct Graph
{
struct Vertex
{
vector<int> o; //, i; //相关出边和入边编号
//int siz, dep, top, dfn; //树链剖分中使用，依次代表子树节点数、深度、所在链的顶端节点、dfs序
};
struct Edge
{
int first, second;
ll cap; //边长、容量，图论算法使用
};
vector<Vertex> v; //点集
vector<Edge> e; //边集
Graph(int n) : v(n) {}
void add(const Edge &ed)
{
if (ed.first == ed.second)
return; //如果有需要请拆点
v[ed.first].o.push_back(e.size());
//v[ed.second].i.push_back(e.size());
e.push_back(ed);
}
//int ch(int u, int i = 0) { return e[v[u].o[i]].second; } //u的第i个孩子节点
//int fa(int u, int i = 0) { return e[v[u].i[i]].first; } //u的第i个父节点
};
struct ISAP : Graph
{
ll flow;
vector<ll> f;
vector<int> h, cur, gap;
ISAP(int n) : Graph(n) {}
void add(Edge ed)
{
Graph::add(ed);
swap(ed.first, ed.second), ed.cap = 0;
Graph::add(ed);
}
ll dfs(int s, int u, int t, ll r)
{
if (r == 0 || u == t)
return r;
ll _f, _r = 0;
for (int &i = cur[u], k; i < v[u].o.size(); ++i)
if (k = v[u].o[i], h[u] == h[e[k].second] + 1)
{
_f = dfs(s, e[k].second, t, min(r - _r, e[k].cap - f[k]));
f[k] += _f, f[k ^ 1] -= _f, _r += _f;
if (_r == r || h[s] >= v.size())
return _r;
}
if (!--gap[h[u]])
h[s] = v.size();
return ++gap[++h[u]], cur[u] = 0, _r;
}
void ask(int s, int t)
{
h.assign(v.size(), 0);
cur.assign(v.size(), 0);
gap.assign(v.size() + 2, 0);
/*
for (deque<int> q(h[t] = gap[t] = 1, t); !q.empty(); q.pop_front()) //优化，加了能快一点
for (int i = 0, u = q.front(), k, to; i < v[u].o.size(); ++i)
if (to = e[v[u].o[i]].second, !h[to])
++gap[h[to] = h[u] + 1], q.push_back(to);
*/
for (f.assign(e.size(), flow = 0); h[s] < v.size();)
flow += dfs(s, s, t, INF);
}
};
int main()
{
int n, m, k, s, t, ss;
scanf("%d%d%d", &n, &m, &k);
ISAP g(2 * n + m + 3);
s = 2 * n + m;
ss = s + 1;
t = ss + 1;
g.add({s, ss, k});
for (int i = 0, tt; i < n; ++i)
{
for (scanf("%d", &tt); tt--;)
{
scanf("%d", &k);
g.add({i, n + n + k - 1, 1});
g.add({n + i, n + n + k - 1, 1});
}
g.add({s, i, 1});
g.add({ss, n + i, 1});
}
for (int i = 0; i < m; ++i)
g.add({n + n + i, t, 1});
g.ask(s, t);
printf("%d", g.flow);
}
Prime Game
计算每个素因子在原序列中出现的位置，然后逐个每个素因子计算对答案的贡献。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e6 + 9;
struct EulerSieve
{
vector<int> p, m; //, phi, mu; //素数序列，最小素因子，欧拉函数，莫比乌斯函数
EulerSieve(int N) : m(N, 0) //, phi(N, 0), mu(N, 0)
{
//phi[1] = mu[1] = 1; //m[1]=0,m[i]==i可判断i是素数
for (long long i = 2, k; i < N; ++i) //防i*p[j]爆int
{
if (!m[i])
p.push_back(m[i] = i); //, phi[i] = i - 1, mu[i] = -1; //i是素数
for (int j = 0; j < p.size() && (k = i * p[j]) < N; ++j)
{
//phi[k] = phi[i] * p[j];
if ((m[k] = p[j]) == m[i])
{
// mu[k] = 0;
break;
}
//phi[k] -= phi[i];
//mu[k] = -mu[i];
}
}
}
void fac(int n, vector<int> &f)
{
if (n < 2)
return;
f.push_back(m[n]);
fac(n / m[n], f);
}
} e(N);
int n;
vector<int> p[N];
int main()
{
scanf("%d", &n);
for (int i = 0, a; i < n; ++i)
{
scanf("%d", &a);
vector<int> fac;
e.fac(a, fac);
fac.resize(unique(fac.begin(), fac.end()) - fac.begin());
for (int j = 0; j < fac.size(); ++j)
p[fac[j]].push_back(i);
}
ll ans = 0;
for (int i = 0; i < e.p.size(); ++i)
{
int a = e.p[i];
if (p[a].empty())
continue;
ans += n * ll(n + 1) >> 1;
for (int i = 0; i <= p[a].size(); ++i)
{
ll len;
if (i == 0)
len = p[a][0];
else if (i == p[a].size())
len = n - p[a].back() - 1;
else
len = p[a][i] - p[a][i - 1] - 1;
ans -= len * (len + 1) >> 1;
}
}
printf("%lld", ans);
}