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CSP-S 模拟赛 17

news 2025/7/15 7:21:55

T1 zzy 的金牌

$1.30 pt s$ 做法

直接爆搜，枚举每个点放几个，用“结构体 + $se t$ ”去重。
代码：

#include <bits/stdc++.h>#define mkpr make_pair
#define fir first
#define sec secondusing namespace std;typedef pair<int, int> pii;
typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;
typedef double db;const int maxn = 3e2 + 7;
const int inf  = 0x3f3f3f3f;
const int mod  = 998244353;int n, K, a[maxn];
namespace Sub1 {struct node {int x[maxn];bool operator < (const node& y) const {for (int i = 1; i < n; ++i)if (x[i] != y.x[i]) return x[i] < y.x[i];return x[n] < y.x[n];}} tmp;int ans;set<node> uni;void dfs(int now) {if (now == K + 1) {for (int i = 1; i <= n; ++i) tmp.x[i] = a[i];sort(tmp.x + 1, tmp.x + n + 1);if (!uni.count(tmp)) {
//				puts("tmp:");
//				for (int i = 1; i <= n; ++i)
//					printf("%d ", tmp.x[i]);
//				puts("");++ans, ans %= mod;uni.insert(tmp);}return ;}for (int i = 1; i <= n; ++i)++a[i], dfs(now + 1), --a[i];}void Main() {dfs(1);printf("%d\n", ans);exit(0);}
}int main() {scanf("%d%d", &n, &K);for (int i = 1; i <= n; ++i) scanf("%d", a + i);if (n <= 7 && K <= 7) Sub1::Main();return 0;
}

$2.60 pt s$ 做法

考虑怎样得到的最终序列是合法的：
设最后得到的序列是 $s_i$ ，我们给 $a$ 和 $s$ 都升序排序（这样就可以避免出现重复的答案），只有 $∀si≥ai\forall s_i \geq a_i$ 且 $∑i=1nsi−∑i=1nai=k\sum_{i = 1}^n s_i - \sum_{i = 1}^n a_i = k$ 才满足条件。
那么我们令 $b_i = s_i - a_i$ ，我们实际上就要求 $b$ 的合法个数，那么就有：

$bi+ai≤bi+1+ai+1b_i + a_i \leq b_{i + 1} + a_{i + 1}$ ；
$∑i=1n=k.\sum_{i = 1}^n = k.$

设 $dp_{i, j, k}$ 表示前 $i$ 个数中， $∑x=1ibx=j\sum_{x = 1}^i b_x = j$ 且 $b_i = k$ 时的方案数，那么枚举上一个数 $b_{i - 1} = p$ 就有转移方程：
$dp_{i, j, k} = \sum_{p = 0}^{j - k} dp_{i - 1, j - k, p}, \space 其中 \space a_{i - 1} + p \leq a_i + k.$
时间复杂度 $O(nk^3)$ 。
代码：

#include <bits/stdc++.h>#define mkpr make_pair
#define fir first
#define sec secondusing namespace std;typedef pair<int, int> pii;
typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;
typedef double db;const int maxn = 3e2 + 7;
const int inf  = 0x3f3f3f3f;
const int mod  = 998244353;int n, m, a[maxn];int dp[maxn][maxn][maxn];
int main() {scanf("%d%d", &n, &m);for (int i = 1; i <= n; ++i) scanf("%d", a + i);sort(a + 1, a + n + 1);for (int i = 0; i <= m; ++i) dp[1][i][i] = 1;for (int i = 2; i <= n; ++i) {for (int j = 0; j <= m; ++j) {for (int k = 0; k <= j; ++k) {for (int p = 0; p <= min(j - k, k + a[i] - a[i - 1]); ++p)(dp[i][j][k] += dp[i - 1][j - k][p]) %= mod;}}}int ans = 0;for (int i = 0; i <= m; ++i)(ans += dp[n][m][i]) %= mod;printf("%d\n", ans);return 0;
}

$3.100 pt s$ 做法

发现最内层转移就是求前缀和，所以直接用 $g_{i, j, k}$ 记录 $dp_{i, j, k}$ 的前缀和，时间复杂度 $O(nk^2)$ 。
代码：

#include <bits/stdc++.h>#define mkpr make_pair
#define fir first
#define sec secondusing namespace std;typedef pair<int, int> pii;
typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;
typedef double db;const int maxn = 3e2 + 7;
const int inf  = 0x3f3f3f3f;
const int mod  = 998244353;int n, m, a[maxn];int dp[maxn][maxn][maxn];
int g[maxn][maxn][maxn];
int main() {scanf("%d%d", &n, &m);for (int i = 1; i <= n; ++i) scanf("%d", a + i);sort(a + 1, a + n + 1);for (int i = 0; i <= m; ++i)dp[1][i][i] = 1, g[1][i][i] = (g[1][i][i - 1] + dp[1][i][i]) % mod;for (int i = 2; i <= n; ++i) {for (int j = 0; j <= m; ++j) {for (int k = 0; k <= j; ++k) {
//				for (int p = 0; p <= min(j - k, k + a[i] - a[i - 1]); ++p)
//					(dp[i][j][k] += dp[i - 1][j - k][p]) %= mod;dp[i][j][k] = g[i - 1][j - k][min(j - k, k + a[i] - a[i - 1])];g[i][j][k] = (g[i][j][k - 1] + dp[i][j][k]) % mod;}}}int ans = 0;for (int i = 0; i <= m; ++i)(ans += dp[n][m][i]) %= mod;printf("%d\n", ans);return 0;
}

T3 作弊

$1.8 pt s$ 做法

直接爆搜将小朋友划分成若干个不相交的区间，然后统计能满足几个小朋友，取其中的 $ma x$ 。
代码：

namespace Sub1 {int b[maxn];  // 枚举每个区间的结束位置int ans;void dfs(int now, int start) {if (start == n + 1) {int L = 1, R = 0, res = 0;for (int i = 1; i <= now; ++i) {R = b[i]; int mx = 0;for (int j = L; j <= R; ++j) mx = max(mx, a[j]);for (int j = L; j <= R; ++j)if (max(a[j], mx) >= l[j] && max(a[j], mx) <= r[j])++res;L = b[i] + 1;}ans = max(res, ans);return ;}for (int i = start; i <= n; ++i) {b[now] = i;dfs(now + 1, i + 1);}}void Main() {dfs(1, 1);printf("%d\n", ans);exit(0);}
}

$2.20 pt s$ 做法

首先可以看出这是一个 $d p$ ，说所以设 $dp_i$ 表示前 $i$ 个同学中最多可以满足的个数。
那么就有转移：
$dp_i = max_{j = 0}^{i - 1} \left \{dp_j + cnt(j + 1, i) \right \}$
其中 $c n t (l, r)$ 表示区间 $[l, r]$ 内赋成最大值后有多少个同学可以被满足。
时间复杂度 $O(n^3)$ ，代码如下：

namespace Sub2 {int dp[107];void Main() {for (int i = 1; i <= n; ++i) {for (int j = 0; j < i; ++j) {int mx = 0, val = 0;for (int k = j + 1; k <= i; ++k) mx = max(mx, a[k]);for (int k = j + 1; k <= i; ++k)if (mx >= l[k] && mx <= r[k]) ++val;dp[i] = max(dp[i], dp[j] + val);}}printf("%d\n", dp[n]);}
}

$3.40 pt s$ 做法

$20 pt s$ 做法的时间瓶颈在求区间最大值和统计区间内被满足的个数。
对于前者，我们可以用 $ST$ 表维护；对于后者我们可以开 $5000$ 颗树状数组， $bit_i.ask(x)$ 统计当最大值为 $i$ 时，前 $x$ 个小朋友有多少个可以被满足。
时间复杂度 $O(n^2log(n))$ 。~~（TMD 维护最大值用的线段树，常数太大没过）~~
代码：

#include <bits/stdc++.h>#define mkpr make_pair
#define fir first
#define sec secondusing namespace std;typedef pair<int, int> pii;
typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;
typedef double db;const int maxn = 1e5 + 7;
const int inf  = 0x3f3f3f3f;int n, a[maxn], L[maxn], R[maxn];int dp[maxn], rt;
int st[maxn][20], lg2[maxn];
void init() {for (int i = 2; i <= n; ++i) lg2[i] = lg2[i >> 1] + 1;for (int i = 1; i <= n; ++i) st[i][0] = a[i];for (int j = 1; j <= lg2[n]; ++j)for (int i = 1; i + (1 << j) - 1 <= n; ++i)st[i][j] = max(st[i][j - 1], st[i + (1 << j - 1)][j - 1]);
}
int ask(int l, int r) {int pw = lg2[r - l + 1];return max(st[l][pw], st[r - (1 << pw) + 1][pw]);
}
struct BIT {int s[5007];void mdf(int p, int x) {for (; p <= n; p += (p & -p)) s[p] += x;}int ask(int p) {int res = 0;for (; p; p -= (p & -p)) res += s[p];return res;}
} bit[5007];int main() {scanf("%d", &n);for (int i = 1; i <= n; ++i) scanf("%d", a + i);	for (int i = 1; i <= n; ++i) scanf("%d%d", L + i, R + i);init();for (int i = 1; i <= n; ++i)for (int j = 1; j <= n; ++j)if (i >= L[j] && i <= R[j]) bit[i].mdf(j, 1);for (int i = 1; i <= n; ++i) {for (int j = 0; j < i; ++j) {int mx = ask(j + 1, i);dp[i] = max(dp[i], dp[j] + bit[mx].ask(i) - bit[mx].ask(j));}}printf("%d\n", dp[n]);return 0;
}

$4.100 pt s$ 做法

发现现在唯一的时间瓶颈就是求 $\left \{ dp_{j - 1} + cst(j, i) \right\}$ ，这和基站选址有点像，可以用线段数优化。
对于 $i$ 而言，我们把条件 $li≤max(l,...,i,...,r)≤ril_i \leq max(l, ..., i, ...,r) \leq r_i$ 变为【 $ma x (l, ..., i)$ 和 $ma x (i, ..., r)$ 都 $≤ri\leq r_i$ 且至少有一个 $≥li\geq l_i$ 】。我们可以用二分求出左右端点的取值范围分别记为 $ll_i, lr_i, rl_i, rr_i$ 。
现在分类讨论：

如果 $lr_i$ 存在，那么对于所有的右端点 $∈[i,rri]\in [i, rr_i]$ 且左端点 $∈[lli,lri]\in [ll_i, lr_i]$ ， $i$ 都有贡献；
如果 $rl_i$ 存在，那么对于所有右端点 $∈[rli,rri]\in [rl_i, rr_i]$ 且左端点 $∈[lli,i]\in [ll_i,i]$ ， $i$ 都有贡献。

代码：

#include <bits/stdc++.h>#define il inline#define mkpr make_pair
#define fir first
#define sec secondusing namespace std;typedef pair<int, int> pii;
typedef unsigned long long ull;
typedef long double ld;
typedef double db;const int maxn = 1e5 + 7;
const int inf  = 0x3f3f3f3f;int n, a[maxn], L[maxn], R[maxn];int dp[maxn], rt;
int st[maxn][20], lg2[maxn];
void init() {for (int i = 2; i <= n; ++i) lg2[i] = lg2[i >> 1] + 1;for (int i = 1; i <= n; ++i) st[i][0] = a[i];for (int j = 1; j <= lg2[n]; ++j)for (int i = 1; i + (1 << j) - 1 <= n; ++i)st[i][j] = max(st[i][j - 1], st[i + (1 << j - 1)][j - 1]);
}
int ask(int l, int r) {int pw = lg2[r - l + 1];return max(st[l][pw], st[r - (1 << pw) + 1][pw]);
}
int ll[maxn], lr[maxn], rl[maxn], rr[maxn];
void init2() {for (int i = 1, l, r; i <= n; ++i) {// 求 ll[i], 即当左端点在 [ll[i], i] 时最大值 <= R[i]l = 1, r = i;while (l <= r) {int mid = l + r >> 1;if (ask(mid, i) <= R[i])ll[i] = mid, r = mid - 1;else l = mid + 1;}// 求 lr[i], 即当左端点 <= lr[i] 时最大值 >= L[i]l = 1, r = i;while (l <= r) {int mid = l + r >> 1;if (ask(mid, i) >= L[i])lr[i] = mid, l = mid + 1;else r = mid - 1;}// 求 rl[i], 即当右端点 >= rl[i] 时最大值 >= L[i]l = i, r = n;while (l <= r) {int mid = l + r >> 1;if (ask(i, mid) >= L[i])rl[i] = mid, r = mid - 1;else l = mid + 1;}// 求 rr[i], 即当右端点在 [i, rr[i]] 时最大值 <= R[i]l = i, r = n;while (l <= r) {int mid = l + r >> 1;if (ask(i, mid) <= R[i])rr[i] = mid, l = mid + 1;else r = mid - 1;}}
}
#define ls (p << 1)
#define rs (p << 1 | 1)
#define mid (l + r >> 1)
struct SegmentTree {int mx[maxn << 2], tg[maxn << 2];il void pushup(int p) {mx[p] = max(mx[ls], mx[rs]);}il void pushdown(int p) {mx[ls] += tg[p], tg[ls] += tg[p];mx[rs] += tg[p], tg[rs] += tg[p];tg[p] = 0;}void mdf(int p, int l, int r, int ql, int qr, int v) {if (ql > qr || qr < l || r < ql) return ;if (ql <= l && r <= qr) {mx[p] += v, tg[p] += v;return ;}pushdown(p);mdf(ls, l, mid, ql, qr, v);mdf(rs, mid + 1, r, ql, qr, v);pushup(p);}int ask(int p, int l, int r, int ql, int qr) {if (ql > qr || qr < l || r < ql) return -inf;if (ql <= l && r <= qr) return mx[p];pushdown(p);return max(ask(ls, l, mid, ql, qr), ask(rs, mid + 1, r, ql, qr));}
} sgt;
struct node {int l, r, v;};
vector<node> v[maxn];
int main() {scanf("%d", &n);for (int i = 1; i <= n; ++i) scanf("%d", a + i);for (int i = 1; i <= n; ++i) scanf("%d%d", L + i, R + i);init(), init2();
//	for (int i = 1; i <= n; ++i)
//		printf("i:%d, ll:%d, lr:%d, rl:%d, rr:%d\n",
//				   i, ll[i], lr[i], rl[i], rr[i]);for (int i = 1; i <= n; ++i) {if (a[i] > R[i] || (!lr[i] && !rl[i])) continue;if (rl[i]) {  // 可能存在 [i, n] 区间内的数的最大值都小于 L[i]v[rl[i]].push_back(node{ll[i], i, 1});v[rr[i] + 1].push_back(node{ll[i], i, -1});}if (lr[i]) {  // 可能存在 [1, i] 区间内的数的最大值都小于 L[i]v[i].push_back(node{ll[i], lr[i], 1});v[rl[i]].push_back(node{ll[i], lr[i], -1});}}for (int i = 1; i <= n; ++i) {for (node p : v[i]) sgt.mdf(1, 1, n, p.l, p.r, p.v);dp[i] = sgt.ask(1, 1, n, 1, i);sgt.mdf(1, 1, n, i + 1, i + 1, dp[i]);
//		printf("dp[%d]:%d\n", i, dp[i]);}printf("%d\n", dp[n]);return 0;
}