20221231 - 模拟赛 | Final Fantasy = xixike = 人间忽晚，山河已秋

不会，二分 / 三分，期望 dp + 高斯消元

# A. 记事本

不会。

$\text{30pts}$ ：建图跑最短路。

$\text{60pts}$ ：分讨模拟。

放个 60 分的代码吧。

	#include <bits/stdc++.h>
	#define pb push_back

	using namespace std;

	namespace IO{
	inline int read(){
	int x = 0, f = 1;
	char ch = getchar();
	while(!isdigit(ch)) {if(ch == '-') f = -1; ch = getchar();}
	while(isdigit(ch)) x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
	return x * f;
	}

	template <typename T> inline void write(T x){
	if(x < 0) putchar('-'), x = -x;
	if(x > 9) write(x / 10);
	putchar(x % 10 + '0');
	}
	}
	using namespace IO;

	const int N = 2010;
	int n, q;
	int a[N], pre[N], nxt[N];
	int mn[N][20];

	inline int qry(int l, int r){
	if(l >= r) return 1e9;
	int k = log2(r - l + 1);
	// cout << l << " " << r << " " << k << " " << r - (1 << k) + 1 << endl;
	return min(mn[l][k], mn[r - (1 << k) + 1][k]);
	}

	signed main(){
	#ifndef ONLINE_JUDGE
	freopen("test.in", "r", stdin);
	freopen("test.out", "w", stdout);
	#endif
	n = read();
	for(int i = 1; i <= n; ++i)
	a[i] = mn[i][0] = read(), pre[i] = nxt[i] = i;
	for(int i = 2; i <= n; ++i)
	if(a[i - 1] < a[i]) pre[i] = pre[i - 1];
	for(int i = n - 1; i >= 1; --i)
	if(a[i + 1] < a[i]) nxt[i] = nxt[i + 1];


	for(int j = 1; j <= 19; ++j)
	for(int i = 1; i + (1 << j) - 1 <= n; ++i)
	mn[i][j] = min(mn[i][j - 1], mn[i + (1 << (j - 1))][j - 1]);
	q = read();
	// for(int i = 44; i <= 58; ++i) cout << a[i] << " ";
	// cout << endl;
	while(q--){
	int r1 = read(), c1 = read(), r2 = read(), c2 = read(), cw = c1, ct = c1;
	// if(abs(r1 - r2) <= 10) cout << "???" << endl;
	int ans = 1e9;
	if(r1 < r2){
	c1 = min(c1, qry(r1, r2));
	// for(int i = r1; i <= r2; ++i) c1 = min(c1, a[i]);
	ans = r2 - r1 + min({abs(c1 - c2), c2 + (c1 != 0), a[r2] - c2 + (c1 != a[r2])});

	for(int i = r1; i <= r2; ++i){
	ct = min(ct, a[i]);
	int flag = (ct != a[i]);
	c1 = a[i];
	c1 = min(c1, qry(i + 1, r2));
	// for(int j = i + 1; j <= r2; ++j) c1 = min(c1, a[j]);
	ans = min(ans, r2 - r1 + flag + abs(c1 - c2));
	}
	for(int i = r2 + 1; i <= n; ++i){
	ct = min(ct, a[i]);
	int flag = (ct != a[i]);
	c1 = a[i];
	c1 = min(c1, qry(r2, i - 1));
	// for(int j = i - 1; j >= r2; --j) c1 = min(c1, a[j]);
	ans = min(ans, i - r1 + i - r2 + abs(c1 - c2) + flag);
	}
	ct = cw;
	for(int i = r1 - 1; i >= 1; --i){
	ct = min(ct, a[i]);
	int flag = (ct != a[i]);
	c1 = a[i];
	c1 = min(c1, qry(i + 1, r2));
	// for(int j = i + 1; j <= r2; ++j) c1 = min(c1, a[j]);
	ans = min(ans, r2 - i + r1 - i + abs(c1 - c2) + flag);
	}
	write(ans), puts("");
	}else{
	c1 = min(c1, qry(r2, r1));
	// for(int i = r2; i <= r1; ++i) c1 = min(c1, a[i]);
	ans = r1 - r2 + min({abs(c1 - c2), c2 + (c1 != 0), a[r2] - c2 + (c1 != a[r2])});
	c1 = cw;
	for(int i = r1; i >= r2; --i){
	ct = min(ct, a[i]);
	int flag = (ct != a[i]);
	c1 = a[i];
	c1 = min(c1, qry(r2, i - 1));
	// for(int j = i - 1; j >= r2; --j) c1 = min(c1, a[j]);
	ans = min(ans, r1 - r2 + flag + abs(c1 - c2));
	}
	for(int i = r2 - 1; i >= 1; --i){
	ct = min(ct, a[i]);
	int flag = (ct != a[i]);
	c1 = a[i];
	c1 = min(c1, qry(i + 1, r2));
	// for(int j = i + 1; j <= r2; ++j) c1 = min(c1, a[j]);
	ans = min(ans, r1 - i + r2 - i + abs(c1 - c2) + flag);
	}
	ct = cw;
	for(int i = r1 + 1; i <= n; ++i){
	ct = min(ct, a[i]);
	int flag = (ct != a[i]);
	c1 = a[i];
	c1 = min(c1, qry(r2, i - 1));
	// for(int j = i - 1; j >= r2; --j) c1 = min(c1, a[j]);
	ans = min(ans, i - r1 + i - r2 + abs(c1 - c2) + flag);
	}
	write(ans), puts("");
	}
	}
	return 0;
	}

# B. 子集

枚举中位数，二分选数个数，手动计算价值。

	#include <bits/stdc++.h>
	#define pb push_back
	#define int long long

	using namespace std;

	namespace IO{
	inline int read(){
	int x = 0, f = 1;
	char ch = getchar();
	while(!isdigit(ch)) {if(ch == '-') f = -1; ch = getchar();}
	while(isdigit(ch)) x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
	return x * f;
	}

	template <typename T> inline void write(T x){
	if(x < 0) putchar('-'), x = -x;
	if(x > 9) write(x / 10);
	putchar(x % 10 + '0');
	}
	}
	using namespace IO;

	const int N = 2e5 + 10;
	int n;
	int a[N], s[N];

	inline double calc(int x, int mid){
	int sum = s[x] - s[x - mid - 1] + s[n] - s[n - mid];
	return 1.0 * sum / (mid + mid + 1) - a[x];
	}

	signed main(){
	#ifndef ONLINE_JUDGE
	freopen("test.in", "r", stdin);
	freopen("test.out", "w", stdout);
	#endif
	n = read();
	for(int i = 1; i <= n; ++i) a[i] = read();
	sort(a + 1, a + 1 + n);
	for(int i = 1; i <= n; ++i) s[i] = s[i - 1] + a[i];
	double ans = 0;

	if(n <= 2000){
	for(int i = 1; i <= n; ++i){
	int l = i - 1, r = n, sum = a[i], cnt = 1;
	while(l >= 1 && r > i){
	sum += a[l] + a[r];
	l--, r--, cnt += 2;
	ans = max(ans, 1.0 * sum / cnt - a[i]);
	}
	}
	}else{
	for(int i = 1; i <= n; ++i){
	int l = 0, r = min(i - 1, n - i);
	while(l <= r){
	int mid = (l + r) >> 1;
	// cout << l << " " << r << " " << midl << " " << midr << endl;
	if(calc(i, mid) > calc(i, mid - 1)) l = mid + 1;
	else r = mid - 1;
	}
	ans = max(ans, calc(i, r));
	// ans = max({ans, calc(i, l), calc(i, r)});
	}
	}
	printf("%.5lf\n", ans);
	return 0;
	}

# C. 子串

先建出 AC 自动机，设 $f_i$ 表示到达 AC 自动机上 $i$ 点时期望再加上多少个点结束。

那么有 $f_{end_i} = 0$ 。

设 $y$ 是 AC 自动机上 $x$ 的子节点，有转移：

$f_x = \frac {1}{26}\sum_{y \in son(x)} f_y + 1$

根据这个式子建出方程组，然后高斯消元即可。

	#include <bits/stdc++.h>
	#define pb push_back

	using namespace std;

	namespace IO{
	inline int read(){
	int x = 0, f = 1;
	char ch = getchar();
	while(!isdigit(ch)) {if(ch == '-') f = -1; ch = getchar();}
	while(isdigit(ch)) x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
	return x * f;
	}

	template <typename T> inline void write(T x){
	if(x < 0) putchar('-'), x = -x;
	if(x > 9) write(x / 10);
	putchar(x % 10 + '0');
	}
	}
	using namespace IO;

	const int N = 200;
	const int mod = 1e9 + 7;
	int T, n, inv26;
	char s[N];

	inline int add(int x) {return x >= mod ? x - mod : x;}
	inline int sub(int x) {return x < 0 ? x + mod : x;}
	inline int mul(int x, int y) {return 1ll * x * y % mod;}

	inline int qpow(int a, int b){
	int res = 1;
	while(b){
	if(b & 1) res = mul(res, a);
	a = mul(a, a), b >>= 1;
	}
	return res;
	}

	int ch[N][26], tot, End[N];
	int fail[N];

	inline void insert(char *s){
	int len = strlen(s), now = 0;
	for(int i = 0; i < len; ++i){
	int c = s[i] - 'a';
	if(!ch[now][c]) ch[now][c] = ++tot;
	now = ch[now][c];
	}
	End[now] = 1;
	}

	inline void build(){
	queue <int> q;
	for(int i = 0; i < 26; ++i)
	if(ch[0][i]) q.push(ch[0][i]);
	while(!q.empty()){
	int x = q.front(); q.pop();
	for(int i = 0; i < 26; ++i){
	if(ch[x][i]) fail[ch[x][i]] = ch[fail[x]][i], q.push(ch[x][i]);
	else ch[x][i] = ch[fail[x]][i];
	}
	}
	}

	int a[N][N];

	inline void Gauss(int n){
	for(int i = 0; i <= n; ++i){
	for(int j = i + 1; j <= n && a[i][i] == 0; ++j)
	if(a[j][i]) swap(a[i], a[j]);
	int inv = qpow(a[i][i], mod - 2);
	for(int j = i + 1; j <= n; ++j)
	for(int k = n + 1; k >= i; --k)
	a[j][k] = sub(a[j][k] - mul(a[i][k], mul(a[j][i], inv)));
	}
	for(int i = n; i >= 0; --i){
	for(int j = i + 1; j <= n; ++j)
	a[i][n + 1] = sub(a[i][n + 1] - mul(a[i][j], a[j][n + 1]));
	a[i][n + 1] = mul(a[i][n + 1], qpow(a[i][i], mod - 2));
	}
	}

	inline void solve(){
	tot = 0;
	memset(a, 0, sizeof(a));
	memset(ch, 0, sizeof(ch));
	memset(End, 0, sizeof(End));
	memset(fail, 0, sizeof(fail));

	n = read();
	for(int i = 1; i <= n; ++i)
	scanf("%s", s), insert(s);
	build();
	for(int i = 0; i <= tot; ++i){
	a[i][i] = 1;
	if(End[i]) continue;
	a[i][tot + 1] = 1;
	for(int j = 0; j < 26; ++j)
	a[i][ch[i][j]] = sub(a[i][ch[i][j]] - inv26);
	}
	Gauss(tot);
	write(a[0][tot + 1]), puts("");
	}

	signed main(){
	#ifndef ONLINE_JUDGE
	freopen("test.in", "r", stdin);
	freopen("test.out", "w", stdout);
	#endif
	T = read(), inv26 = qpow(26, mod - 2);
	while(T--) solve();
	return 0;
	}

# A. 记事本

# B. 子集

# C. 子串

20221211

20230117