2A. Politics

因为编号为 $1$ 的人一定不会离开，那么最后留下的人一定要和编号为 $1$ 的人的所有参数都一致，所以计数即可。

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/hash_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif
typedef long long ll;
typedef pair < int, int > PII;
typedef int itn;
mt19937 RND_MAKER (chrono :: steady_clock :: now ().time_since_epoch ().count ());
inline ll randomly (const ll l, const ll r) {return (RND_MAKER () ^ (1ull << 63)) % (r - l + 1) + l;}
//#define int long long
const double pi = acos (-1);
//__gnu_pbds :: tree < Key, Mapped, Cmp_Fn = std :: less < Key >, Tag = rb_tree_tag, Node_Upadte = null_tree_node_update, Allocator = std :: allocator < char > > ;
//__gnu_pbds :: tree < PPS, __gnu_pbds :: null_type, less < PPS >, __gnu_pbds :: rb_tree_tag, __gnu_pbds :: tree_order_statistics_node_update > tr;
inline int read () {
	int x = 0, f = 0;
	char c = getchar ();
	for ( ; c < '0' || c > '9' ; c = getchar ()) f |= (c == '-');
	for ( ; c >= '0' && c <= '9' ; c = getchar ()) x = (x << 1) + (x << 3) + (c & 15);
	return !f ? x : -x;
}
char s[105][105];
int n, k;
signed main () {
	int _ = read ();
	while (_ --) {
		n = read (), k = read ();
		for (int i = 1;i <= n; ++ i) scanf ("%s", s[i] + 1);
		int ans = 1;
		for (int i = 2;i <= n; ++ i) {
			int ok = 1;
			for (int j = 1;j <= k; ++ j) {
				if (s[i][j] != s[1][j]) ok = 0;
			}
			ans += ok;
		} 
		printf ("%d\n", ans);
	}
	return 0;
}

2B. Indivisible

我们可以写一个 $O (n!)$ 的暴力打出表来，然后就发现无解条件是 $n > 1$ 且 $n mod 2 = 1$ 。

然后分成两种情况：

如果 $n = 1$ ，那么答案为 $[1]$ 。
否则答案为 $[2, 1, 4, 3, \dots, 2 n, 2 n - 1]$ 。

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/hash_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif
typedef long long ll;
typedef pair < int, int > PII;
typedef int itn;
mt19937 RND_MAKER (chrono :: steady_clock :: now ().time_since_epoch ().count ());
inline ll randomly (const ll l, const ll r) {return (RND_MAKER () ^ (1ull << 63)) % (r - l + 1) + l;}
//#define int long long
const double pi = acos (-1);
//__gnu_pbds :: tree < Key, Mapped, Cmp_Fn = std :: less < Key >, Tag = rb_tree_tag, Node_Upadte = null_tree_node_update, Allocator = std :: allocator < char > > ;
//__gnu_pbds :: tree < PPS, __gnu_pbds :: null_type, less < PPS >, __gnu_pbds :: rb_tree_tag, __gnu_pbds :: tree_order_statistics_node_update > tr;
inline int read () {
	int x = 0, f = 0;
	char c = getchar ();
	for ( ; c < '0' || c > '9' ; c = getchar ()) f |= (c == '-');
	for ( ; c >= '0' && c <= '9' ; c = getchar ()) x = (x << 1) + (x << 3) + (c & 15);
	return !f ? x : -x;
}
int a[105], _, n, sum[105];
signed main () {
	_ = read ();
	while (_ --) {
		n = read ();
		if (n == 1) {
			printf ("1\n");
			continue;
		}
		if (n & 1) {
			printf ("-1\n");
			continue;
		}
		for (int i = 1, j = 2;i <= n; i += 2, j += 2) a[i] = j;
		for (int i = 2;i <= n; i += 2) a[i] = a[i - 1] - 1;
		for (int i = 1;i <= n; ++ i) printf ("%d ", a[i]);
		printf ("\n");
	}
	return 0;
}

2C/1A. Almost Increasing Subsequence

我们首先考虑长度为 $r - l + 1$ 不合法在哪里。

我们要把满足 $a_{i} \geq a_{i + 1} \geq a_{i + 2}$ 全部删掉，只需要将把 $l \leq i \leq r - 2$ 并且 $a_{i} \geq a_{i + 1} \geq a_{i + 2}$ 的 $i$ 全部删掉。

直接前缀和即可。

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/hash_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif
typedef long long ll;
typedef pair < int, int > PII;
typedef int itn;
mt19937 RND_MAKER (chrono :: steady_clock :: now ().time_since_epoch ().count ());
inline ll randomly (const ll l, const ll r) {return (RND_MAKER () ^ (1ull << 63)) % (r - l + 1) + l;}
//#define int long long
const double pi = acos (-1);
//__gnu_pbds :: tree < Key, Mapped, Cmp_Fn = std :: less < Key >, Tag = rb_tree_tag, Node_Upadte = null_tree_node_update, Allocator = std :: allocator < char > > ;
//__gnu_pbds :: tree < PPS, __gnu_pbds :: null_type, less < PPS >, __gnu_pbds :: rb_tree_tag, __gnu_pbds :: tree_order_statistics_node_update > tr;
inline int read () {
	int x = 0, f = 0;
	char c = getchar ();
	for ( ; c < '0' || c > '9' ; c = getchar ()) f |= (c == '-');
	for ( ; c >= '0' && c <= '9' ; c = getchar ()) x = (x << 1) + (x << 3) + (c & 15);
	return !f ? x : -x;
}
int n, q, a[200005], s[200005];
signed main () {
	n = read (), q = read ();
	for (int i = 1;i <= n; ++ i) a[i] = read ();
	for (int i = 1;i <= n; ++ i) {
		s[i] = s[i - 1];
		if (a[i] >= a[i + 1] && a[i + 1] >= a[i + 2] && i + 2 <= n) s[i] ++;
	}
	while (q --) {
		int l = read (), r = read (), ans = r - l + 1;
		int L = l, R = r - 2;
		if (L <= R) {
			int d = s[R] - s[L - 1];
			ans -= d;
		}
		printf ("%d\n", ans);
	}
	return 0;
}

2D/1B. Fish Graph

我们考虑将每一个连通块分开处理，对于每一个连通块，取出其中任意一个生成树，拎出来每一个非树边，然后就构成了一个环，枚举每一个点（是点 $u$ ），判断是不是一个 Fish Garph 即可。

代码非常难写，有 114514 个细节，需要注意。

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/hash_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif
typedef long long ll;
typedef pair < int, int > PII;
typedef int itn;
mt19937 RND_MAKER (chrono :: steady_clock :: now ().time_since_epoch ().count ());
inline ll randomly (const ll l, const ll r) {return (RND_MAKER () ^ (1ull << 63)) % (r - l + 1) + l;}
//#define int long long
const double pi = acos (-1);
//__gnu_pbds :: tree < Key, Mapped, Cmp_Fn = std :: less < Key >, Tag = rb_tree_tag, Node_Upadte = null_tree_node_update, Allocator = std :: allocator < char > > ;
//__gnu_pbds :: tree < PPS, __gnu_pbds :: null_type, less < PPS >, __gnu_pbds :: rb_tree_tag, __gnu_pbds :: tree_order_statistics_node_update > tr;
inline int read () {
	int x = 0, f = 0;
	char c = getchar ();
	for ( ; c < '0' || c > '9' ; c = getchar ()) f |= (c == '-');
	for ( ; c >= '0' && c <= '9' ; c = getchar ()) x = (x << 1) + (x << 3) + (c & 15);
	return !f ? x : -x;
}
const int N = 2005;
struct LCAns {
	struct edge {int to, nxt;} e[N << 1];
	int head[N], eid;
	inline void add_edge (int u, int v) {e[eid] = {v, head[u]}; head[u] = eid ++;}
	int dp[N][17], d[N], lg[N];
	int n, root;
	inline void dfs (int u) {
		for (int i = head[u]; ~i ; i = e[i].nxt) {
			int v = e[i].to;
			if (v == dp[u][0]) continue; 
			dp[v][0] = u; d[v] = d[u] + 1; dfs (v);
		}
	}
	inline int LCA (int u, int v) {
		if (d[u] < d[v]) swap (u, v);
		for (int j = lg[d[u] - d[v]];j >= 0; -- j) {if (d[dp[u][j]] >= d[v]) u = dp[u][j];}
		if (u == v) return u;
		for (int j = lg[d[u]];j >= 0; -- j) {if (dp[u][j] != dp[v][j]) {u = dp[u][j], v = dp[v][j];}}
		return dp[u][0];
	}
	inline void perf () {
		for (int i = 0;i < N; ++ i) head[i] = -1;
		eid = 0;
	}
	inline void pre () {
		lg[0] = -1;
		for (int i = 1;i <= n; ++ i) dp[i][0] = 0, d[i] = 0;
		for (int i = 1;i <= n; ++ i) lg[i] = lg[i / 2] + 1;
		d[root] = 1; dfs (root);
		for (int j = 1; (1 << j) < n; ++ j) for (int i = 1;i <= n; ++ i) {
			dp[i][j] = dp[dp[i][j - 1]][j - 1];
		}
	}
} tr; 
int ed[N][2], n, m, vis[N], vis2[N], fl[N];
struct DSU {
	int fa[N];
	inline void init (int X) {
		for (int i = 1;i <= X; ++ i) fa[i] = i;
	}
	inline int find (int x) {
		return fa[x] == x ? x : fa[x] = find (fa[x]);
	}
	inline void merge (int x, int y) {
		fa[find (x)] = find (y);
	}
	inline int query (int x, int y) {
		return find (x) == find (y);
	}
} awa, qwq;
vector < pair < int, int > > th;
vector < int > G2[N], G[N];
signed main () {
	int _ = read ();
	while (_ --) {
		n = read (), m = read ();
		for (int i = 1;i <= n; ++ i) vis[i] = 0;
		for (int i = 1;i <= m; ++ i) vis2[i] = 0;
		awa.init (n);
		for (int i = 1;i <= m; ++ i) {
			ed[i][0] = read (), ed[i][1] = read ();
			awa.merge (ed[i][0], ed[i][1]);
			G[ed[i][0]].push_back (ed[i][1]);
			G[ed[i][1]].push_back (ed[i][0]);
		}
		int ok = 0;
		for (int i = 1;i <= n; ++ i) {
			if (vis[i]) continue;
			tr.n = n, tr.root = i;
			tr.perf ();
			vector < int > _edge, ex;
			_edge.clear ();
			ex.clear ();
			qwq.init (n);
			for (int j = 1;j <= m; ++ j) {
				if (!awa.query (i, ed[j][0]) || !awa.query (i, ed[j][1])) continue;
				if (qwq.query (ed[j][0], ed[j][1])) {
					ex.push_back (j);
					continue;
				}
				tr.add_edge (ed[j][0], ed[j][1]);
				tr.add_edge (ed[j][1], ed[j][0]);
				G2[ed[j][0]].push_back (ed[j][1]);
				G2[ed[j][1]].push_back (ed[j][0]);
				vis2[j] = 1;
				vis[ed[j][0]] = vis[ed[j][1]] = 1;
				_edge.push_back (j);
				qwq.merge (ed[j][0], ed[j][1]);
			}
			tr.pre ();
			for (int E : ex) {
				int U = ed[E][0], V = ed[E][1];
				int Z = tr.LCA (U, V);
				for (int j = 1;j <= n; ++ j) fl[j] = 1;
				vector < int > vertex;
				vertex.clear ();
				int tot = 1;
				for (int j = U; j != Z; j = tr.dp[j][0]) vertex.push_back (j), tot ++;
				for (int j = V; j != Z; j = tr.dp[j][0]) vertex.push_back (j), tot ++;
				vertex.push_back (Z);
				for (int u : vertex) fl[u] = 0;
				for (int u : vertex) {
					th.clear ();
					for (int v : G[u]) {
						if (fl[v]) th.push_back (make_pair (u, v));
					}
					if (th.size () >= 2) {
						tot += 2;
						printf ("YES\n");
						ok = 1;
						printf ("%d\n", tot);
						for (int j = U; j != Z; j = tr.dp[j][0]) printf ("%d %d\n", j, tr.dp[j][0]);
						for (int j = V; j != Z; j = tr.dp[j][0]) printf ("%d %d\n", j, tr.dp[j][0]);
						printf ("%d %d\n", U, V);
						printf ("%d %d\n", th[0].first, th[0].second);
						printf ("%d %d\n", th[1].first, th[1].second);
						break;
					}
				}
				if (ok) break;
			}
			for (int j = 1;j <= n; ++ j) G2[j].clear ();
			if (ok) break;
		}
		if (!ok) printf ("NO\n");
		for (int i = 1;i <= n; ++ i) G[i].clear ();
	}
	return 0;
}

2E/1C. Similar Polynomials

首先我们设：

A (x) = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{d - 1} x^{d - 1} + a_{d} x^{d}

B (x) = b_{0} + b_{1} x + b_{2} x^{2} + \dots + b_{d - 1} x^{d - 1} + b_{d} x^{d}

我们要让 $B (x - s) = A (x)$ ，我们观察 $d$ 次项系数：

b_{d} (x - s)^{d} = trash + b_{d} \times x^{d}

发现永远不会变。

然后我们要让 $d - 1$ 次项系数相等，发现：

发现会对 $d - 1$ 次项系数产生贡献的只有 $b_{d - 1} (x - s)^{d - 1}$ 和 $b_{d} (x - s)^{d}$ 。

然后就爆算：

[x^{d - 1}] B (x - s) = b_{d - 1} - b_{d} s d = a_{d - 1}

由此可得：

s = \frac{b_{d - 1} - a_{d - 1}}{b_{d} d}

然后就是拉格朗日插值求出 $[x^{d}] A (x), [x^{d}] B (x), [x^{d - 1}] A (x), [x^{d - 1}] B (x)$ 。

拉格朗日插值的式子：

f (x) = \sum_{i = 0}^{d} y_{i} \prod_{j \neq i} \frac{x - x_{j}}{x_{i} - x_{j}}

常数项就不用理会了，因为乘起来的贡献为 $1$ ，i. e.：

[x^{d}] f (x) = \sum_{i = 0}^{d} y_{i} \prod_{j \neq i} \frac{1}{x_{i} - x_{j}} = \sum_{i = 0}^{d} y_{i} \prod_{j \neq i} \frac{1}{i - j} = \sum_{i = 0}^{d} y_{i} \frac{1}{i! \times (d - i)! \times (- 1)^{d - i}}

同理：

[x^{k - 1}] \prod_{i = 1}^{k} (x - x_{i}) = - \sum_{i = 1}^{k} x_{k}

然后就是爆算：

[x^{d - 1}] f (x) = - \sum_{i = 0}^{d} y_{i} \prod_{j \neq i} \frac{d (d + 1) / 2 - i}{x_{i} - x_{j}} = - \sum_{i = 0}^{d} y_{i} \prod_{j \neq i} \frac{d (d + 1) / 2 - i}{i - j}

= - \sum_{i = 0}^{d} y_{i} \frac{d (d + 1) / 2 - i}{i! \times (d - i)! \times (- 1)^{d - i}}

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/hash_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif
typedef long long ll;
typedef pair < int, int > PII;
typedef int itn;
mt19937 RND_MAKER (chrono :: steady_clock :: now ().time_since_epoch ().count ());
inline ll randomly (const ll l, const ll r) {return (RND_MAKER () ^ (1ull << 63)) % (r - l + 1) + l;}
#define int long long
const double pi = acos (-1);
//__gnu_pbds :: tree < Key, Mapped, Cmp_Fn = std :: less < Key >, Tag = rb_tree_tag, Node_Upadte = null_tree_node_update, Allocator = std :: allocator < char > > ;
//__gnu_pbds :: tree < PPS, __gnu_pbds :: null_type, less < PPS >, __gnu_pbds :: rb_tree_tag, __gnu_pbds :: tree_order_statistics_node_update > tr;
inline int read () {
	int x = 0, f = 0;
	char c = getchar ();
	for ( ; c < '0' || c > '9' ; c = getchar ()) f |= (c == '-');
	for ( ; c >= '0' && c <= '9' ; c = getchar ()) x = (x << 1) + (x << 3) + (c & 15);
	return !f ? x : -x;
}
const int mod = 1e9 + 7;
const int N = 2.5e6 + 5;
inline int pow_mod (int a, int b, int p) {
	int res = 1;
	while (b) {
		if (b & 1) res = res * a % p;
		b >>= 1, a = a * a % p;
	}
	return res;
}
int d, A[N], B[N], coefA[2], coefB[2], fac[N], ifac[N];
signed main () {
	d = read ();
	fac[0] = 1;
	for (int i = 1;i < N; ++ i) fac[i] = fac[i - 1] * i % mod;
	ifac[N - 1] = pow_mod (fac[N - 1], mod - 2, mod);
	for (int i = N - 2;i >= 0; -- i) ifac[i] = ifac[i + 1] * (i + 1) % mod;
	for (int i = 0;i <= d; ++ i) A[i] = read ();
	for (int i = 0;i <= d; ++ i) B[i] = read ();
	for (int i = 0;i <= d; ++ i) {
		int dx = ifac[i] * ifac[d - i] % mod;
		if ((d - i) & 1) dx = mod - dx;
		if (dx >= mod) dx -= mod;
		dx = dx * A[i] % mod;
		coefA[1] = (coefA[1] + dx) % mod;
	}
	for (int i = 0;i <= d; ++ i) {
		int dx = ifac[i] * ifac[d - i] % mod;
		if ((d - i) & 1) dx = mod - dx;
		if (dx >= mod) dx -= mod;
		dx = dx * B[i] % mod;
		coefB[1] = (coefB[1] + dx) % mod;
	}
	for (int i = 0;i <= d; ++ i) {
		int dx = ifac[i] * ifac[d - i] % mod;
		if ((d - i) & 1) dx = mod - dx;
		if (dx >= mod) dx -= mod;
		dx = dx * A[i] % mod;
		int coef = d * (d + 1) / 2 - i;
		coef %= mod;
		dx = dx * coef % mod;
		coefA[0] = (coefA[0] + dx) % mod;
	}
	for (int i = 0;i <= d; ++ i) {
		int dx = ifac[i] * ifac[d - i] % mod;
		if ((d - i) & 1) dx = mod - dx;
		if (dx >= mod) dx -= mod;
		dx = dx * B[i] % mod;
		int coef = d * (d + 1) / 2 - i;
		coef %= mod;
		dx = dx * coef % mod;
		coefB[0] = (coefB[0] + dx) % mod;
	}
	coefA[0] = mod - coefA[0], coefA[0] %= mod;
	coefB[0] = mod - coefB[0], coefB[0] %= mod;
	int ans = coefB[1] * d % mod;
	ans = pow_mod (ans, mod - 2, mod);
	int coef = coefB[0] - coefA[0];
	coef = (coef % mod + mod) % mod;
	ans = ans * coef % mod;
	printf ("%lld\n", ans);
	return 0;	
}

2F/1D. Toy Machine

对于中间的方块的编号设为 $m i d$ 。

对于所有 $k < m i d$ 的点，发现只需要不断的做 LDRU 即可，最后再做一遍 $L$ ，大致是将最左边的移走。

对于 $k = m i d$ 的点，答案就是 DL，为什么自己想。

对于 $k > m i d$ 的点，这个就很妙了，我们分步处理：

确定一个数据：15 10。

Step $1$ ：

中间的 g 有点烦人，先移走：

Step $2$ ：

可以用中间的黑块卡住一个方块，剩下的整体向下，操作序列为 ULDL。

Step $3$ ：

在保留顺序的情况下将方块卡到右半边：

因为要保留顺序，所以先整体向右：

然后就是按 URDR 的顺序一个个移到右半边，正确性显然。

Step $4$ ：

出答案。

将填充好的上边移动到左边，然后就按照 $k < m i d$ 的做就行了：

最后就是这个效果：

代码比较简单，但是需要自己计算。

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/hash_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif
typedef long long ll;
typedef pair < int, int > PII;
typedef int itn;
mt19937 RND_MAKER (chrono :: steady_clock :: now ().time_since_epoch ().count ());
inline ll randomly (const ll l, const ll r) {return (RND_MAKER () ^ (1ull << 63)) % (r - l + 1) + l;}
//#define int long long
const double pi = acos (-1);
//__gnu_pbds :: tree < Key, Mapped, Cmp_Fn = std :: less < Key >, Tag = rb_tree_tag, Node_Upadte = null_tree_node_update, Allocator = std :: allocator < char > > ;
//__gnu_pbds :: tree < PPS, __gnu_pbds :: null_type, less < PPS >, __gnu_pbds :: rb_tree_tag, __gnu_pbds :: tree_order_statistics_node_update > tr;
inline int read () {
	int x = 0, f = 0;
	char c = getchar ();
	for ( ; c < '0' || c > '9' ; c = getchar ()) f |= (c == '-');
	for ( ; c >= '0' && c <= '9' ; c = getchar ()) x = (x << 1) + (x << 3) + (c & 15);
	return !f ? x : -x;
}
signed main () {
	int n = read (), k = read (), num = n - 2, mid = (num + 1) / 2;
	if (k < mid) {
		for (int i = 1;i < k; ++ i) printf ("LDRU");
		printf ("L\n");
	}
	else if (k == mid) printf ("DL\n");
	else {
		printf ("DL");
		for (int i = mid + 1;i <= k; ++ i) printf ("ULDL");
		printf ("R");
		for (int i = 1;i <= num - (mid + 1) + 1; ++ i) printf ("URDR");
		printf ("L");
		for (int i = 1;i < mid; ++ i) printf ("LDRU");
		printf ("L\n");
	}
	return 0;
}

posted @ 2023-04-30 18:31 CountingGroup 阅读(322) 评论(2) 编辑收藏举报

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