Atcoder Codefestival Exhibition/Team Relay/Tournament Round 简要题解

Exhibition

Awkward

考虑容斥，至少某些限制不满足。

把不满足的边画出来，发现是若干条不相交路径，DP即可。

#include <bits/stdc++.h>
#define xx first
#define yy second
#define mp make_pair
#define pb push_back
#define mset(x, y) memset(x, y, sizeof x)
#define mcpy(x, y) memcpy(x, y, sizeof x)
using namespace std;typedef long long LL;
typedef pair <int, int> pii;inline int Read()
{int x = 0, f = 1, c = getchar();for (; !isdigit(c); c = getchar())if (c == '-')f = -1;for (;  isdigit(c); c = getchar())x = x * 10 + c - '0';return x * f;
}const int MAXN = 2005;
const int mod = 1e9 + 7;int n, siz[MAXN], g[MAXN][3], f[MAXN][MAXN][3];
vector <int> adj[MAXN];inline void DFS(int x)
{siz[x] = 1, f[x][1][0] = 1;for (auto y : adj[x]){DFS(y), mcpy(g, f[x]), mset(f[x], 0);for (int dx = 0; dx <= 2; dx ++)for (int dy = 0; dy <= 2; dy ++)for (int dz = 0; dz <= 1; dz ++)if (!dz || (dx < 2 && dy < 2))for (int i = 1; i <= siz[x]; i ++)for (int j = 1; j <= siz[y]; j ++){int cur = 1;if (dz){cur = mod - 1;if (!dx && !dy)cur = (cur << 1) % mod;if (dx && dy)cur = cur >> 1;}f[x][i + j - dz][dx + dz] = (1LL * g[i][dx] * f[y][j][dy] % mod * cur + f[x][i + j - dz][dx + dz]) % mod;}siz[x] += siz[y];}
}int main()
{
#ifdef wxh010910freopen("data.in", "r", stdin);
#endifn = Read();for (int i = 2; i <= n; i ++)adj[Read()].pb(i);DFS(1);int ret = 0;for (int i = 0, cur = 1; i <= n; i ++, cur = 1LL * cur * i % mod)for (int j = 0; j <= 2; j ++)ret = (1LL * f[1][i][j] * cur + ret) % mod;return printf("%d\n", ret), 0;
}

Increment and Swap

从左往右考虑，假设当前考虑第ii个数，记costv=v+∑i−1j=1[aj>v]cost_v = v + \sum_{j=1}^{i-1}[a_j > v]，可以证明每次贪心取costvcost_v最小的最优，操作后将costj(1≤j<ai)cost_j(1\le j 加一，然后线段树维护。

证明：考虑反证，如果存在一个更优的v′v'更优，显然v′<vv' 。

接下来归纳，首先i=ni=n时成立，如果i=k+1i=k+1时成立，那么我们只需要证明i=ki=k时成立即可。

注意到costv′>costvcost_{v'}>cost_v，而在之后的操作中，仍然有costi>costv(1≤i<v)cost_i > cost_v(1\le i （考虑每次操作的影响），而之后贪心是对的，所以这一步贪心也是对的。

#include <bits/stdc++.h>
#define xx first
#define yy second
#define mp make_pair
#define pb push_back
#define mset(x, y) memset(x, y, sizeof x)
#define mcpy(x, y) memcpy(x, y, sizeof x)
using namespace std;typedef long long LL;
typedef pair <int, int> pii;inline int Read()
{int x = 0, f = 1, c = getchar();for (; !isdigit(c); c = getchar())if (c == '-')f = -1;for (;  isdigit(c); c = getchar())x = x * 10 + c - '0';return x * f;
}const int MAXN = 200005;int n, m, a[MAXN], b[MAXN], tag[MAXN << 2];
pii val[MAXN << 2];
LL ans;inline void Build(int x, int l, int r)
{val[x] = mp(b[l], l);if (l == r)return ;int mid = l + r >> 1;Build(x << 1, l, mid), Build(x << 1 | 1, mid + 1, r);
}inline void Modify(int x, int l, int r, int ql, int qr)
{if (l == ql && r == qr)return (void)(val[x].xx ++, tag[x] ++);if (tag[x])val[x << 1].xx += tag[x], tag[x << 1] += tag[x], val[x << 1 | 1].xx += tag[x], tag[x << 1 | 1] += tag[x], tag[x] = 0;int mid = l + r >> 1;if (qr <= mid)Modify(x << 1, l, mid, ql, qr);else if (ql > mid)Modify(x << 1 | 1, mid + 1, r, ql, qr);elseModify(x << 1, l, mid, ql, mid), Modify(x << 1 | 1, mid + 1, r, mid + 1, qr);val[x] = min(val[x << 1], val[x << 1 | 1]);
}inline pii Query(int x, int l, int r, int ql, int qr)
{if (l == ql && r == qr)return val[x];if (tag[x])val[x << 1].xx += tag[x], tag[x << 1] += tag[x], val[x << 1 | 1].xx += tag[x], tag[x << 1 | 1] += tag[x], tag[x] = 0;int mid = l + r >> 1;if (qr <= mid)return Query(x << 1, l, mid, ql, qr);if (ql > mid)return Query(x << 1 | 1, mid + 1, r, ql, qr);return min(Query(x << 1, l, mid, ql, mid), Query(x << 1 | 1, mid + 1, r, mid + 1, qr));
}int main()
{
#ifdef wxh010910freopen("data.in", "r", stdin);
#endifn = Read();for (int i = 1; i <= n; i ++)a[i] = b[i] = Read(), ans -= a[i];sort(b + 1, b + n + 1), m = unique(b + 1, b + n + 1) - b - 1, Build(1, 1, m);for (int i = 1; i <= n; i ++){pii val = Query(1, 1, m, lower_bound(b + 1, b + m + 1, a[i]) - b, m);ans += val.xx;if (val.yy > 1)Modify(1, 1, m, 1, val.yy - 1);}return printf("%lld\n", ans), 0;
}

Tournament Round

Colorful MST

把颜色相同的缩起来，跑MST，两个并查集维护原图连通性和颜色连通性即可。

#include <bits/stdc++.h>
#define xx first
#define yy second
#define mp make_pair
#define pb push_back
#define mset(x, y) memset(x, y, sizeof x)
#define mcpy(x, y) memcpy(x, y, sizeof x)
using namespace std;typedef long long LL;
typedef pair <int, int> pii;inline int Read()
{int x = 0, f = 1, c = getchar();for (; !isdigit(c); c = getchar())if (c == '-')f = -1;for (;  isdigit(c); c = getchar())x = x * 10 + c - '0';return x * f;
}const int MAXN = 100005;struct Edge
{int u, v, w;bool operator < (const Edge &b) const{return w < b.w;}
} e[MAXN];int n, m, c, f[MAXN], g[MAXN], col[MAXN];
LL ret;inline int FindF(int x)
{while (x ^ f[x])x = f[x] = f[f[x]];return x;
}inline int FindG(int x)
{while (x ^ g[x])x = g[x] = g[g[x]];return x;
}int main()
{
#ifdef wxh010910freopen("data.in", "r", stdin);
#endifn = Read(), m = Read(), c = Read() - 1;for (int i = 1; i <= n; i ++)col[i] = Read(), f[i] = i;for (int i = 1; i <= c + 1; i ++)g[i] = i;for (int i = 1; i <= m; i ++)e[i].u = Read(), e[i].v = Read(), e[i].w = Read();sort(e + 1, e + m + 1);for (int i = 1; i <= m; i ++){if (!c)break;int u = FindF(e[i].u), v = FindF(e[i].v), cu = col[u], cv = col[v];if (u == v)continue;f[u] = v, col[v] = cv ? cv : cu;if (cu && cv){cu = FindG(cu), cv = FindG(cv);if (cu == cv)continue;g[cu] = cv, c --, ret += e[i].w;}elsec --, ret += e[i].w;}if (c)ret = -1;return printf("%lld\n", ret), 0;
}

Many Swaps Sorting

注意到只用11和n−1n-1就能完成了，冒泡排序。

#include <bits/stdc++.h>
#define xx first
#define yy second
#define mp make_pair
#define pb push_back
#define mset(x, y) memset(x, y, sizeof x)
#define mcpy(x, y) memcpy(x, y, sizeof x)
using namespace std;typedef long long LL;
typedef pair <int, int> pii;inline int Read()
{int x = 0, f = 1, c = getchar();for (; !isdigit(c); c = getchar())if (c == '-')f = -1;for (;  isdigit(c); c = getchar())x = x * 10 + c - '0';return x * f;
}const int MAXN = 205;vector <int> ans;
int n, a[MAXN];inline void Solve(int x)
{ans.pb(x);for (int i = x; i < n; i ++)swap(a[i], a[i - x]);
}int main()
{
#ifdef wxh010910freopen("data.in", "r", stdin);
#endifn = Read();for (int i = 0; i < n; i ++)a[i] = Read();while (!is_sorted(a, a + n))if (a[0] && a[0] < a[n - 1])Solve(n - 1);elseSolve(1);printf("%d\n", ans.size());for (int i = 0; i < ans.size(); i ++)printf("%d\n", ans[i]);return 0;
}

Paired Parentheses

只需要满足：s1=t1,s2n=t2ns_1 = t_1, s_{2n} = t_{2n}，并且相等的对数和不相等的对数都是偶数即可，构造很显然。

线段树维护。

#include <bits/stdc++.h>
#define xx first
#define yy second
#define mp make_pair
#define pb push_back
#define mset(x, y) memset(x, y, sizeof x)
#define mcpy(x, y) memcpy(x, y, sizeof x)
using namespace std;typedef long long LL;
typedef pair <int, int> pii;inline int Read()
{int x = 0, f = 1, c = getchar();for (; !isdigit(c); c = getchar())if (c == '-')f = -1;for (;  isdigit(c); c = getchar())x = x * 10 + c - '0';return x * f;
}const int MAXN = 200005;LL val_a[MAXN << 2], val_b[MAXN << 2];
int n, m, a[MAXN], b[MAXN];inline void Build(int x, int l, int r)
{if (l == r)return (void)(val_a[x] = a[l], val_b[x] = b[l]);int mid = l + r >> 1;Build(x << 1, l, mid), Build(x << 1 | 1, mid + 1, r);val_a[x] = max(val_a[x << 1] + val_a[x << 1 | 1], val_b[x << 1] + val_b[x << 1 | 1]);val_b[x] = max(val_b[x << 1] + val_a[x << 1 | 1], val_a[x << 1] + val_b[x << 1 | 1]);
}inline void Modify(int x, int l, int r, int p)
{if (l == r)return (void)(val_a[x] = a[l], val_b[x] = b[l]);int mid = l + r >> 1;if (p <= mid)Modify(x << 1, l, mid, p);elseModify(x << 1 | 1, mid + 1, r, p);val_a[x] = max(val_a[x << 1] + val_a[x << 1 | 1], val_b[x << 1] + val_b[x << 1 | 1]);val_b[x] = max(val_b[x << 1] + val_a[x << 1 | 1], val_a[x << 1] + val_b[x << 1 | 1]);
}int main()
{
#ifdef wxh010910freopen("data.in", "r", stdin);
#endifn = Read(), m = Read();for (int i = 1; i <= n << 1; i ++)a[i] = Read();for (int i = 1; i <= n << 1; i ++)b[i] = Read();if (n == 1){while (m --){int p = Read(), x = Read(), y = Read();a[p] = x, b[p] = y;printf("%d\n", a[1] + a[n << 1]);}return 0;}Build(1, 2, (n << 1) - 1);while (m --){int p = Read(), x = Read(), y = Read();a[p] = x, b[p] = y;if (p > 1 && p < n << 1)Modify(1, 2, (n << 1) - 1, p);printf("%lld\n", a[1] + a[n << 1] + val_a[1]);}return 0;
}

Ancient Tree Record

用相邻两个点的权值可以求出这条边的权值，两边大小相等的边最多一条，最后暴力求。

#include <bits/stdc++.h>
#define xx first
#define yy second
#define mp make_pair
#define pb push_back
#define mset(x, y) memset(x, y, sizeof x)
#define mcpy(x, y) memcpy(x, y, sizeof x)
using namespace std;typedef long long LL;
typedef pair <int, int> pii;inline LL Read()
{LL x = 0; int f = 1, c = getchar();for (; !isdigit(c); c = getchar())if (c == '-')f = -1;for (;  isdigit(c); c = getchar())x = x * 10 + c - '0';return x * f;
}const int MAXN = 100005;int n, e_cnt = 1, p[MAXN << 1], v[MAXN << 1], hed[MAXN], siz[MAXN];
LL cur, a[MAXN], val[MAXN];inline void Addedge(int x, int y)
{e_cnt ++, p[e_cnt] = y, v[e_cnt] = hed[x], hed[x] = e_cnt;e_cnt ++, p[e_cnt] = x, v[e_cnt] = hed[y], hed[y] = e_cnt;
}inline void DFS(int x, int par)
{siz[x] = 1;for (int i = hed[x]; i; i = v[i])if (p[i] ^ par){DFS(p[i], x);if (siz[p[i]] << 1 ^ n)val[i >> 1] = (a[x] - a[p[i]]) / ((siz[p[i]] << 1) - n);siz[x] += siz[p[i]];}
}inline void Cal(int x, int par)
{siz[x] = 1;for (int i = hed[x]; i; i = v[i])if (p[i] ^ par)Cal(p[i], x), cur += siz[p[i]] * val[i >> 1], siz[x] += siz[p[i]];
}int main()
{
#ifdef wxh010910freopen("data.in", "r", stdin);
#endifn = Read();for (int i = 1; i < n; i ++)Addedge(Read(), Read());for (int i = 1; i <= n; i ++)a[i] = Read();DFS(1, 0);for (int i = 1; i < n; i ++)if (!val[i]){int u = p[i << 1], v = p[i << 1 | 1];Cal(u, v), Cal(v, u);val[i] = (a[u] - cur) / siz[v];}for (int i = 1; i < n; i ++)printf("%lld\n", val[i]);return 0;
}

Black Cats Deployment

将边从大到小加入，启发式合并。

#include <bits/stdc++.h>
#define xx first
#define yy second
#define mp make_pair
#define pb push_back
#define mset(x, y) memset(x, y, sizeof x)
#define mcpy(x, y) memcpy(x, y, sizeof x)
using namespace std;typedef long long LL;
typedef pair <int, int> pii;inline int Read()
{int x = 0, f = 1, c = getchar();for (; !isdigit(c); c = getchar())if (c == '-')f = -1;for (;  isdigit(c); c = getchar())x = x * 10 + c - '0';return x * f;
}const int MAXN = 100005;struct Edge
{int u, v, w;bool operator < (const Edge &b) const{return w > b.w;}
} e[MAXN];LL ans[MAXN], tag[MAXN];
vector <int> g[MAXN];
int n, f[MAXN];inline int Find(int x)
{while (x ^ f[x])x = f[x] = f[f[x]];return x;
}inline void Merge(int x, int y, int w)
{x = Find(x), y = Find(y);if (g[x].size() < g[y].size())swap(x, y);for (auto i : g[y])ans[i] += tag[y] - tag[x], ans[i] += 1LL * (g[x].size() - g[y].size()) * w;tag[x] += 1LL * g[y].size() * w;for (auto i : g[y])g[x].pb(i);f[y] = x;
}int main()
{
#ifdef wxh010910freopen("data.in", "r", stdin);
#endifn = Read();for (int i = 1; i <= n; i ++)f[i] = i, g[i].pb(i);for (int i = 1; i < n; i ++)e[i].u = Read(), e[i].v = Read(), e[i].w = Read();sort(e + 1, e + n);for (int i = 1; i < n; i ++)Merge(e[i].u, e[i].v, e[i].w);for (int i = 1; i <= n; i ++)printf("%lld\n", ans[i] + tag[Find(i)]);return 0;
}

Unicyclic Graph Counting

将环缩起来，定义其prufer序列是不删去环上点，并且最后一个点在环上，那么不难发现环上的点出现了d−2d-2次，非环上的点出现了d−1d-1次。

f(i,j,0/1)f(i,j,0/1)表示前ii个点，有jj个点在环上，是否prufer序列的最后一个点出现过的方案数。

#include <bits/stdc++.h>
#define xx first
#define yy second
#define mp make_pair
#define pb push_back
#define mset(x, y) memset(x, y, sizeof x)
#define mcpy(x, y) memcpy(x, y, sizeof x)
using namespace std;typedef long long LL;
typedef pair <int, int> pii;inline int Read()
{int x = 0, f = 1, c = getchar();for (; !isdigit(c); c = getchar())if (c == '-')f = -1;for (;  isdigit(c); c = getchar())x = x * 10 + c - '0';return x * f;
}const int MAXN = 305;
const int mod = 1e9 + 7;int n, a[MAXN], fac[MAXN], inv[MAXN], f[MAXN][MAXN][2];int main()
{
#ifdef wxh010910freopen("data.in", "r", stdin);
#endifn = Read(), fac[0] = fac[1] = inv[0] = inv[1] = 1;for (int i = 2; i <= n; i ++)fac[i] = 1LL * fac[i - 1] * i % mod, inv[i] = 1LL * (mod - mod / i) * inv[mod % i] % mod;bool sam = true;for (int i = 1; i <= n; i ++)inv[i] = 1LL * inv[i - 1] * inv[i] % mod, a[i] = Read(), sam &= a[i] == 2;if (sam)return printf("%d\n", 1LL * fac[n - 1] * inv[2] % mod), 0;f[0][0][0] = 1;for (int i = 0; i < n; i ++)for (int j = 0; j <= i; j ++)for (int k = 0; k < 2; k ++)if (f[i][j][k]){f[i + 1][j][k] = (f[i + 1][j][k] + f[i][j][k]) % mod;if (a[i + 1] >= 2)f[i + 1][j + 1][k] = (1LL * f[i][j][k] * (a[i + 1] - 1) + f[i + 1][j + 1][k]) % mod;if (a[i + 1] >= 3 && !k)f[i + 1][j + 1][1] = (1LL * f[i][j][k] * (a[i + 1] - 1) * (a[i + 1] - 2) + f[i + 1][j + 1][1]) % mod;}int ans = 0, cur = 1;for (int i = 1; i <= n; i ++)cur = 1LL * cur * inv[a[i] - 1] % mod;for (int i = 3; i < n; i ++)if (f[n][i][1])ans = (1LL * fac[i - 1] % mod * f[n][i][1] % mod * fac[n - i - 1] % mod * inv[2] + ans) % mod;return printf("%d\n", 1LL * ans * cur % mod), 0;
}

Team Relay

Kaiden

分类讨论一下。

#include <bits/stdc++.h>
#define xx first
#define yy second
#define mp make_pair
#define pb push_back
#define mset(x, y) memset(x, y, sizeof x)
#define mcpy(x, y) memcpy(x, y, sizeof x)
using namespace std;typedef long long LL;
typedef pair <int, int> pii;inline LL Read()
{LL x = 0; int f = 1, c = getchar();for (; !isdigit(c); c = getchar())if (c == '-')f = -1;for (;  isdigit(c); c = getchar())x = x * 10 + c - '0';return x * f;
}int main()
{
#ifdef wxh010910freopen("data.in", "r", stdin);
#endifLL k = Read(), a = Read(), b = Read();if (a >= k)return puts("1"), 0;if (a <= b)return puts("-1"), 0;return printf("%lld\n", 2 * ((k - b - 1) / (a - b)) + 1), 0;
}

Evergrowing Tree

暴力。

#include <bits/stdc++.h>
#define xx first
#define yy second
#define mp make_pair
#define pb push_back
#define mset(x, y) memset(x, y, sizeof x)
#define mcpy(x, y) memcpy(x, y, sizeof x)
using namespace std;typedef long long LL;
typedef pair <int, int> pii;inline int Read()
{int x = 0, f = 1, c = getchar();for (; !isdigit(c); c = getchar())if (c == '-')f = -1;for (;  isdigit(c); c = getchar())x = x * 10 + c - '0';return x * f;
}int main()
{
#ifdef wxh010910freopen("data.in", "r", stdin);
#endifint n = Read(), m = Read();if (n == 1){while (m --)printf("%d\n", min(Read(), Read()));return 0;}while (m --){int a = Read(), b = Read();while (a ^ b){if (a < b)swap(a, b);a = (a + n - 2) / n;}printf("%d\n", a);}return 0;
}

Garden

二分。

#include <bits/stdc++.h>
#define xx first
#define yy second
#define mp make_pair
#define pb push_back
#define mset(x, y) memset(x, y, sizeof x)
#define mcpy(x, y) memcpy(x, y, sizeof x)
using namespace std;typedef long long LL;
typedef pair <int, int> pii;inline LL Read()
{LL x = 0; int f = 1, c = getchar();for (; !isdigit(c); c = getchar())if (c == '-')f = -1;for (;  isdigit(c); c = getchar())x = x * 10 + c - '0';return x * f;
}const int MAXN = 100005;int n, m, d[MAXN];
LL w[MAXN];inline bool Chk(LL x)
{LL cur = 0;for (int i = 1; i <= n; i ++){if (x >= w[i])cur += (x - w[i]) / d[i] + 1;if (cur >= m)return true;}return false;
}int main()
{
#ifdef wxh010910freopen("data.in", "r", stdin);
#endifn = Read(), m = Read();for (int i = 1; i <= n; i ++)w[i] = Read(), d[i] = Read();LL l = 0, r = 1LL << 61, ret = r;while (l <= r){LL mid = l + r >> 1;if (Chk(mid))ret = mid, r = mid - 1;elsel = mid + 1;}return printf("%lld\n", ret), 0;
}

Shock

贪心往原先size大的加，最后变成两个连通块连成完全图。

#include <bits/stdc++.h>
#define xx first
#define yy second
#define mp make_pair
#define pb push_back
#define mset(x, y) memset(x, y, sizeof x)
#define mcpy(x, y) memcpy(x, y, sizeof x)
using namespace std;typedef long long LL;
typedef pair <int, int> pii;inline int Read()
{int x = 0, f = 1, c = getchar();for (; !isdigit(c); c = getchar())if (c == '-')f = -1;for (;  isdigit(c); c = getchar())x = x * 10 + c - '0';return x * f;
}const int MAXN = 100005;int n, m, f[MAXN], siz[MAXN];inline int Find(int x)
{while (x ^ f[x])x = f[x] = f[f[x]];return x;
}inline void Merge(int x, int y)
{x = Find(x), y = Find(y);if (x == y)return ;siz[x] += siz[y], f[y] = x;
}int main()
{
#ifdef wxh010910freopen("data.in", "r", stdin);
#endifn = Read(), m = Read();for (int i = 1; i <= n; i ++)f[i] = i, siz[i] = 1;for (int i = 1; i <= m; i ++)Merge(Read(), Read());for (int i = 1; i <= n; i ++)if (Find(i) == i && (Find(i) ^ Find(1)) && (Find(i) ^ Find(2)))if (siz[Find(1)] >= siz[Find(2)])siz[Find(1)] += siz[i];elsesiz[Find(2)] += siz[i];LL ret = 1LL * siz[Find(1)] * (siz[Find(1)] - 1) / 2 + 1LL * siz[Find(2)] * (siz[Find(2)] - 1) / 2;return printf("%lld\n", ret - m), 0;
}

White and Blue

WW+B≥P100\frac{W}{W+B}\ge \frac{P}{100}，化一下式子，按照每个人的影响排序，贪心选取。

#include <bits/stdc++.h>
#define xx first
#define yy second
#define mp make_pair
#define pb push_back
#define mset(x, y) memset(x, y, sizeof x)
#define mcpy(x, y) memcpy(x, y, sizeof x)
using namespace std;typedef long long LL;
typedef pair <int, int> pii;inline int Read()
{int x = 0, f = 1, c = getchar();for (; !isdigit(c); c = getchar())if (c == '-')f = -1;for (;  isdigit(c); c = getchar())x = x * 10 + c - '0';return x * f;
}const int MAXN = 100005;int n, p;
LL cur;struct Node
{int w, b;bool operator < (const Node &c) const{return 1LL * w * (100 - p) + 1LL * b * p > 1LL * c.w * (100 - p) + 1LL * c.b * p;}
} a[MAXN];int main()
{
#ifdef wxh010910freopen("data.in", "r", stdin);
#endifn = Read(), p = Read();for (int i = 1; i <= n; i ++)a[i].w = Read(), a[i].b = Read(), cur -= a[i].b;sort(a + 1, a + n + 1);cur *= p;for (int i = 1; i <= n; i ++)if ((cur += 1LL * a[i].w * (100 - p) + 1LL * a[i].b * p) >= 0)return printf("%d\n", i), 0;
}

Capture

让sumr−suml−1−xr+xl−1sum_r - sum_{l-1} - x_r + x_{l-1}最大，扫一遍。

#include <bits/stdc++.h>
#define xx first
#define yy second
#define mp make_pair
#define pb push_back
#define mset(x, y) memset(x, y, sizeof x)
#define mcpy(x, y) memcpy(x, y, sizeof x)
using namespace std;typedef long long LL;
typedef pair <int, int> pii;inline LL Read()
{LL x = 0; int f = 1, c = getchar();for (; !isdigit(c); c = getchar())if (c == '-')f = -1;for (;  isdigit(c); c = getchar())x = x * 10 + c - '0';return x * f;
}LL x, s, ans, cur = LLONG_MIN;
int n;int main()
{
#ifdef wxh010910freopen("data.in", "r", stdin);
#endifn = Read();for (int i = 1; i <= n; i ++)x = Read(), cur = max(cur, x - s), s += Read(), ans = max(ans, cur + s - x);return printf("%lld\n", ans), 0;
}

Coinage

不妨设s+t≤t+ss+t\le t+s，那么ss越多越优。

#include <bits/stdc++.h>
#define xx first
#define yy second
#define mp make_pair
#define pb push_back
#define mset(x, y) memset(x, y, sizeof x)
#define mcpy(x, y) memcpy(x, y, sizeof x)
using namespace std;typedef long long LL;
typedef pair <int, int> pii;inline int Read()
{int x = 0, f = 1, c = getchar();for (; !isdigit(c); c = getchar())if (c == '-')f = -1;for (;  isdigit(c); c = getchar())x = x * 10 + c - '0';return x * f;
}int main()
{
#ifdef wxh010910freopen("data.in", "r", stdin);
#endifstring s, t, ans = "";int n = Read(), x = 0;cin >> s >> t;if (s + t > t + s)swap(s, t);for (int i = 1; i * s.length() <= n; i ++)if ((n - i * s.length()) % t.length() == 0)x = i;for (int i = 1; i <= x; i ++)ans += s;while (ans.length() ^ n)ans += t;cout << ans << endl;return 0;
}

Akashic Records

根号以下的暴力枚举次幂，根号以上的独立，贪心即可。

#include <bits/stdc++.h>
#define xx first
#define yy second
#define mp make_pair
#define pb push_back
#define mset(x, y) memset(x, y, sizeof x)
#define mcpy(x, y) memcpy(x, y, sizeof x)
using namespace std;typedef long long LL;
typedef pair <int, int> pii;inline int Read()
{int x = 0, f = 1, c = getchar();for (; !isdigit(c); c = getchar())if (c == '-')f = -1;for (;  isdigit(c); c = getchar())x = x * 10 + c - '0';return x * f;
}const int MAXN = 305;
const int p[7] = {2, 3, 5, 7, 11, 13, 17};
const int lim[7] = {9, 6, 4, 3, 3, 3, 3};int n, ans, cur[7], w[MAXN], val[MAXN], a[MAXN][8];inline void Solve()
{int now = 0;for (int i = 1; i <= 300; i ++)w[i] = 0;for (int i = 1; i <= n; i ++){bool flg = true;for (int j = 0; j < 7 && flg; j ++)flg &= cur[j] >= a[i][j];if (flg){if (a[i][7] ^ 1)w[a[i][7]] += val[i];elsenow += val[i];}}for (int i = 1; i <= 300; i ++)now += max(w[i], 0);ans = max(ans, now);
}inline void DFS(int x)
{if (x == 7)Solve();elsefor (cur[x] = 0; cur[x] < lim[x]; cur[x] ++)DFS(x + 1);
}int main()
{
#ifdef wxh010910freopen("data.in", "r", stdin);
#endifn = Read();for (int i = 1; i <= n; i ++){a[i][7] = Read(), val[i] = Read();for (int j = 0; j < 7; j ++)while (a[i][7] % p[j] == 0)a[i][7] /= p[j], a[i][j] ++;}DFS(0);return printf("%d\n", ans), 0;
}

Nice to Meet You

补集转化，求两个人不能到达同一个地点的概率。

记两个人能到达的地点集合为AA，BB，显然AA，BB之间没有边，其概率为fAfB12EA+EBf_A f_B \frac{1}{2^{E_A + E_B}}，fSf_S表示只用SS中的点11或者22能到达的点集为SS的概率，ESE_S表示一个端点在SS内，另一个不在的边数。

计算ff可以枚举子集容斥。

#include <bits/stdc++.h>
#define xx first
#define yy second
#define mp make_pair
#define pb push_back
#define mset(x, y) memset(x, y, sizeof x)
#define mcpy(x, y) memcpy(x, y, sizeof x)
using namespace std;typedef long long LL;
typedef pair <int, int> pii;inline int Read()
{int x = 0, f = 1, c = getchar();for (; !isdigit(c); c = getchar())if (c == '-')f = -1;for (;  isdigit(c); c = getchar())x = x * 10 + c - '0';return x * f;
}const int MAXN = 20;
const int MAXM = 32770;
const int mod = 1e9 + 7;int n, m, bin[MAXN * MAXN], e[MAXM], f[MAXM];inline int F(int x, int y)
{return e[x | y] - e[x] - e[y];
}int main()
{
#ifdef wxh010910freopen("data.in", "r", stdin);
#endifn = Read(), m = Read(), bin[0] = 1;for (int i = 1; i <= m; i ++)bin[i] = 1LL * bin[i - 1] * (mod + 1) / 2 % mod;for (int i = 1; i <= m; i ++)for (int j = 0, x = Read() - 1, y = Read() - 1; j < 1 << n; j ++)if ((j >> x & 1) && (j >> y & 1))e[j] ++;for (int i = 0; i < 1 << n; i ++)if ((i & 1) && !(i & 2)){f[i] = 1;for (int j = i - 1; j; j = j - 1 & i)if (j & 1)f[i] = (f[i] - 1LL * f[j] * bin[F(j, j ^ i)] % mod + mod) % mod;}else if ((i & 2) && !(i & 1)){f[i] = 1;for (int j = i - 1; j; j = j - 1 & i)if (j & 2)f[i] = (f[i] - 1LL * f[j] * bin[F(j, j ^ i)] % mod + mod) % mod;}for (int i = 0; i < 1 << n; i ++)f[i] = 1LL * f[i] * bin[F(i, (1 << n) - 1 ^ i)] % mod;int ans = 1;for (int i = 0; i < 1 << n; i ++)if ((i & 1) && !(i & 2))for (int j = (1 << n) - 1 - i; j; j = j - 1 & (1 << n) - 1 - i)if ((j & 2) && !(j & 1))if (!F(i, j))ans = (ans - 1LL * f[i] * f[j] % mod + mod) % mod;for (int i = 1; i <= m; i ++)ans = (ans << 1) % mod;return printf("%d\n", ans), 0;
}

Indifferent

打表发现第ii大的贡献为2n−i2n−1\frac{2n-i}{2n-1}。

证明可以归纳。

#include <bits/stdc++.h>
#define xx first
#define yy second
#define mp make_pair
#define pb push_back
#define mset(x, y) memset(x, y, sizeof x)
#define mcpy(x, y) memcpy(x, y, sizeof x)
using namespace std;typedef long long LL;
typedef pair <int, int> pii;inline int Read()
{int x = 0, f = 1, c = getchar();for (; !isdigit(c); c = getchar())if (c == '-')f = -1;for (;  isdigit(c); c = getchar())x = x * 10 + c - '0';return x * f;
}const int MAXN = 200005;int n, a[MAXN];int main()
{
#ifdef wxh010910freopen("data.in", "r", stdin);
#endifn = Read();for (int i = 0; i < n << 1; i ++)a[i] = Read();sort(a, a + (n << 1));LL ans = 0;for (int i = 0; i < n << 1; i ++)ans += 1LL * a[i] * i;return printf("%.9lf\n", (double)ans / ((n << 1) - 1)), 0;
}

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Atcoder Codefestival Exhibition/Team Relay/Tournament Round 简要题解

Exhibition

Awkward

Increment and Swap

Tournament Round

Colorful MST

Many Swaps Sorting

Paired Parentheses

Ancient Tree Record

Black Cats Deployment

Unicyclic Graph Counting

Team Relay

Kaiden

Evergrowing Tree

Garden

Shock

White and Blue

Capture

Coinage

Akashic Records

Nice to Meet You

Indifferent

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