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📂模拟赛记录

🔖模拟赛题解

2023-10-16 23:00阅读: 54评论: 3推荐: 1

CSP模拟赛记录

2023.10.15

A. 超速

题面有点绕差评

直接二分

可爱的code捏

 // Author: xiaruize
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define mms(arr, n) memset(arr, n, sizeof(arr))
#define rep(i, a, n) for (int i = (a); i <= (n); ++i)
#define per(i, n, a) for (int i = (n); i >= (a); --i)
int max(int a, int b) {
    if (a > b)
        return a;
    return b;
}
int min(int a, int b) {
    if (a < b)
        return a;
    return b;
}
void chkmx(int &x, int y) {
    if (y > x)
        x = y;
}
void chkmi(int &x, int y) {
    if (y < x)
        x = y;
}
const int INF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1000000007;
const int N = 2e5 + 10;
 
// bool st;
int n;
int v[N], l[N];
int m;
int a[N], f[N];
int q;
int lim;
// bool en;
 
bool check(int x) {
    double t = 0;
    rep(i, 1, n) { t = t + (double)l[i] / (v[i] + a[x]); }
    return t < lim;
}
 
void solve() {
    int s, t;
    cin >> s >> t;
    lim = t - s;
    int l = 0, r = m;
    while (l < r) {
        int mid = l + r >> 1;
        if (check(mid))
            r = mid;
        else
            l = mid + 1;
    }
    cout << f[r] << endl;
}
 
signed main() {
    freopen("speed.in", "r", stdin);
    freopen("speed.out", "w", stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int testcase = 1;
    cin >> n;
    rep(i, 1, n) cin >> v[i];
    rep(i, 1, n) cin >> l[i];
    cin >> m;
    a[m + 1] = 1e9;
    rep(i, 1, m - 1) cin >> a[i];
    rep(i, 1, m) cin >> f[i];
    cin >> testcase;
    while (testcase--) solve();
    return 0;
}

B. 对常规的斗争

二次差分板子题挺显然的

可爱的code捏

 // Author: xiaruize
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define mms(arr, n) memset(arr, n, sizeof(arr))
#define rep(i, a, n) for (int i = (a); i <= (n); ++i)
#define per(i, n, a) for (int i = (n); i >= (a); --i)
int max(int a, int b) {
    if (a > b)
        return a;
    return b;
}
int min(int a, int b) {
    if (a < b)
        return a;
    return b;
}
void chkmx(int &x, int y) {
    if (y > x)
        x = y;
}
void chkmi(int &x, int y) {
    if (y < x)
        x = y;
}
const int INF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1000000007;
const int N = 2e5 + 10;
 
// bool st;
int n;
int a[N], s[N];
map<int, int> mp;
// bool en;
 
void solve() {
    cin >> n;
    rep(i, 1, n) {
        cin >> a[i];
        mp[a[i]] = 0;
    }
    rep(i, 1, n) {
        int t = i - mp[a[i]];
        s[1]++;
        s[t + 1]--;
        // cerr << t << endl;
        t = n - i + 1;
        s[t + 1]--;
        s[n - mp[a[i]] + 2]++;
        // rep(i, 0, 10)
        // {
        //     cerr << s[i] << ' ';
        // }
        // cerr << endl;
        mp[a[i]] = i;
    }
    rep(i, 1, n) s[i] += s[i - 1];
    rep(i, 1, n) s[i] += s[i - 1];
    rep(i, 1, n) { cout << s[i] << ' '; }
}
 
signed main() {
    freopen("fight.in", "r", stdin);
    freopen("fight.out", "w", stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int testcase = 1;
    // cin >> testcase;
    while (testcase--) solve();
    return 0;
}

C. 海报

Part 1 不考虑查询

我们不考虑环，假设是在一条链上，那么一个很显然的dp状态为 $dp_{i,j}$ 表示考虑到第 $i$ 个人，从 $i$ 开始往前连续 $j$ 个人举海报

然后考虑环上的情况，我们发现：可以把最前面有多少个连续的人举海报这个东西扔到状态里面即增加一维 $k$ ，表示最前面有多少个连续的人举海报

时间复杂度 $\mathcal{O}(n)$ 空间复杂度 $\mathcal{O}(n)$

Part 2 考虑查询

暴力修改重构，时间复杂度 $\mathcal{O(nq)}$ ，显然不能通过

考虑用线段树维护 dp 数组，具体来说，线段树上每个节点存一个 $dp_{i,j}$ 表示当前节点表示的区间中，前 $i$ 个人和后 $j$ 个人都举海报，且第 $i+1$ 个人和倒数第 $j+1$ 个人不举海报的最大价值，合并的时候暴力转移即可

特别注意要特判区间全部选中的情况

时间复杂度 $\mathcal{O}(nlogn)$ 空间复杂度 $\mathcal{O}(n)$

可爱的code捏

 // Author: xiaruize
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define mms(arr, n) memset(arr, n, sizeof(arr))
#define rep(i, a, n) for (int i = (a); i <= (n); ++i)
#define per(i, n, a) for (int i = (n); i >= (a); --i)
int max(int a, int b) {
    if (a > b)
        return a;
    return b;
}
int min(int a, int b) {
    if (a < b)
        return a;
    return b;
}
void chkmx(int &x, int y) {
    if (y > x)
        x = y;
}
void chkmi(int &x, int y) {
    if (y < x)
        x = y;
}
const int INF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1000000007;
const int N = 4e4 + 10;
 
// bool st;
int n;
int a[N];
int q;
// bool en;
 
struct segment_tree {
#define ls p << 1
#define rs p << 1 | 1
 
    struct node {
        int dp[4][4];
    } tr[N << 2];
 
    void pushup(int p, int l, int r) {
        memset(tr[p].dp, 0, sizeof(tr[p].dp));
        int mid = l + r >> 1;
        rep(i, 0, 3) {
            if (i > mid - l + 1)
                break;
            rep(j, 0, 3) {
                if (j > r - mid)
                    break;
                rep(k, 0, 3) {
                    if (k > mid - l + 1)
                        break;
                    rep(d, 0, 3) {
                        if (d > r - mid)
                            break;
                        if (k + d > 3)
                            continue;
                        if (i == mid - l + 1 && j == r - mid) {
                            if (i + j > 3)
                                continue;
                            tr[p].dp[i + j][i + j] =
                                max(tr[p].dp[i + j][i + j], tr[ls].dp[i][i] + tr[rs].dp[j][j]);
                        } else if (i == mid - l + 1) {
                            if (i + d > 3)
                                continue;
                            tr[p].dp[i + d][j] = max(tr[p].dp[i + d][j], tr[ls].dp[i][i] + tr[rs].dp[d][j]);
                        } else if (j == r - mid) {
                            if (j + k > 3)
                                continue;
                            tr[p].dp[i][j + k] = max(tr[p].dp[i][j + k], tr[ls].dp[i][k] + tr[rs].dp[j][j]);
                        } else
                            tr[p].dp[i][j] = max(tr[p].dp[i][j], tr[ls].dp[i][k] + tr[rs].dp[d][j]);
                    }
                }
            }
        }
    }
 
    void build(int p, int l, int r) {
        if (l == r) {
            tr[p].dp[1][1] = a[l];
            return;
        }
        int mid = l + r >> 1;
        build(ls, l, mid);
        build(rs, mid + 1, r);
        pushup(p, l, r);
        // cerr << p << ' ' << l << ' ' << r << endl;
        // rep(i, 0, 3)
        // {
        // 	rep(j, 0, 3)
        // 	{
        // 		cerr << tr[p].dp[i][j] << ' ';
        // 	}
        // 	cerr << endl;
        // }
    }
 
    void update(int p, int l, int r, int pos, int val) {
        if (l == r) {
            tr[p].dp[1][1] = val;
            return;
        }
        int mid = l + r >> 1;
        if (mid >= pos)
            update(ls, l, mid, pos, val);
        else
            update(rs, mid + 1, r, pos, val);
        pushup(p, l, r);
        // cerr << p << ' ' << l << ' ' << r << endl;
        // rep(i, 0, 3)
        // {
        // 	rep(j, 0, 3)
        // 	{
        // 		cerr << tr[p].dp[i][j] << ' ';
        // 	}
        // 	cerr << endl;
        // }
    }
 
    int getans() {
        int res = 0;
        rep(i, 0, 3) {
            rep(j, 0, 3) {
                if (i + j <= 3) {
                    res = max(res, tr[1].dp[i][j]);
                }
            }
        }
        return res;
    }
} seg;
 
void solve() {
    cin >> n;
    rep(i, 1, n) { cin >> a[i]; }
    seg.build(1, 1, n);
    cout << seg.getans() << endl;
    cin >> q;
    while (q--) {
        int x, v;
        cin >> x >> v;
        seg.update(1, 1, n, x, v);
        cout << seg.getans() << endl;
    }
}
 
signed main() {
    freopen("poster.in", "r", stdin);
    freopen("poster.out", "w", stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int testcase = 1;
    // cin >> testcase;
    while (testcase--) solve();
    return 0;
}

D. 机器人锦标赛

题面还是比较烦但是其实暴力加就可以遇到 $1$ 显然就不用继续操作了

可爱的code捏

 // Author: xiaruize
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define mms(arr, n) memset(arr, n, sizeof(arr))
#define rep(i, a, n) for (int i = (a); i <= (n); ++i)
#define per(i, n, a) for (int i = (n); i >= (a); --i)
int max(int a, int b) {
    if (a > b)
        return a;
    return b;
}
int min(int a, int b) {
    if (a < b)
        return a;
    return b;
}
void chkmx(int &x, int y) {
    if (y > x)
        x = y;
}
void chkmi(int &x, int y) {
    if (y < x)
        x = y;
}
const int INF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1000000007;
const int N = 3e5 + 10;
 
// bool st;
struct node {
    int op, a, b;
};
vector<node> op[N];
vector<int> g[N], p[N], fa[N];
bool vis[N];
int n, m, s, res, f[N];
// bool en;
 
void upd(int x, int y) {
    if (vis[x])
        return;
    g[x][y] = 1;
    for (int v = y, t;; v = t) {
        if (v == m * 2 - 2)
            break;
        t = fa[x][v];
        if (op[x][t - m].op == 1)
            g[x][t] = g[x][op[x][t - m].a] & g[x][op[x][t - m].b];
        else
            g[x][t] = g[x][op[x][t - m].a] | g[x][op[x][t - m].b];
    }
    if (g[x][m * 2 - 2]) {
        vis[x] = true;
        res++;
    }
}
 
void solve() {
    cin >> m >> n >> s;
    rep(i, 1, n) {
        g[i].resize(m * 2);
        fa[i].resize(m * 2);
        op[i].resize(m);
        rep(j, 0, m - 2) {
            cin >> op[i][j].a >> op[i][j].b >> op[i][j].op;
            op[i][j].a--;
            op[i][j].b--;
            fa[i][op[i][j].a] = fa[i][op[i][j].b] = j + m;
        }
    }
    rep(i, 0, m - 1) p[i].resize(n);
    rep(i, 1, n) {
        rep(j, 0, m - 1) {
            int x;
            cin >> x;
            p[j][x] = i;
        }
    }
    rep(i, 0, m - 1) {
        while (res < s && f[i] < n) upd(p[i][f[i]++], i);
    }
    rep(i, 0, m - 1) cout << f[i] << ' ';
}
 
signed main() {
    freopen("robot.in", "r", stdin);
    freopen("robot.out", "w", stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int testcase = 1;
    // cin >> testcase;
    while (testcase--) solve();
    return 0;
}

2023.10.15比赛总结

300 分没挂，最后一题不打暴力有点可惜

2023.10.16

A. 魔力子串

直接vector 扔 map里面没什么好说的

警示后人: 能用map就不要哈希

可爱的code捏

 #include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define double long double
#define mms(arr, n) memset(arr, n, sizeof(arr))
#define rep(i, a, n) for (int i = (a); i <= (n); ++i)
#define per(i, n, a) for (int i = (n); i >= (a); --i)
int max(int a, int b) {
    if (a > b)
        return a;
    return b;
}
int min(int a, int b) {
    if (a < b)
        return a;
    return b;
}
const int INF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1000000007;
const int N = 2e5 + 10;
 
// bool st;
int n;
string s;
int cnt[52];
map<vector<int>, int> mp;
map<char, int> chg;
int tot = 0;
// bool en;
 
int ch(char c) {
    if (chg.count(c))
        return chg[c];
    return chg[c] = tot++;
}
 
void solve() {
    cin >> n >> s;
    rep(i, 0, n - 1) ch(s[i]);
    vector<int> tmp(tot);
    mp[tmp]++;
    rep(i, 0, n - 1) {
        cnt[ch(s[i])]++;
        vector<int> tmp(tot);
        rep(i, 0, tot - 2) {
            tmp[i] = cnt[i + 1] - cnt[i];
            // cerr << tmp[i] << ' ';
        }
        // cerr << endl;
        mp[tmp]++;
    }
    int res = 0;
    for (auto v : mp) {
        // cerr << v.second << endl;
        res += v.second * (v.second - 1) / 2;
        res %= MOD;
    }
    cout << res << endl;
}
 
signed main() {
    freopen("magic.in", "r", stdin);
    freopen("magic.out", "w", stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int testcase = 1;
    // cin >> testcase;
    while (testcase--) solve();
    return 0;
}

B. 吃树

结论题

当正好存在 $\frac{n}{k}$ 个节点的子树大小为 $k$ 的倍数时, $k$ 作为块的大小是合法的
对于每种合法的块的大小，有且仅有 $1$ 种构造方案

可爱的code捏

 #include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define double long double
#define mms(arr, n) memset(arr, n, sizeof(arr))
#define rep(i, a, n) for (int i = (a); i <= (n); ++i)
#define per(i, n, a) for (int i = (n); i >= (a); --i)
int max(int a, int b)
{
	if (a > b)
		return a;
	return b;
}
int min(int a, int b)
{
	if (a < b)
		return a;
	return b;
}
const int INF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1000000007;
const int N = 1e6 + 10;
 
// bool st;
int n;
vector<int> g[N];
int siz[N], cnt[N];
// bool en;
 
void dfs(int x, int fa)
{
	siz[x] = 1;
	for (auto v : g[x])
	{
		if (v == fa)
			continue;
		dfs(v, x);
		siz[x] += siz[v];
	}
	cnt[siz[x]]++;
}
 
int res = 0;
 
void calc(int x)
{
	int c = 0;
	for (int i = x; i <= n; i += x)
	{
		c += cnt[i];
	}
	if (c == n / x)
	{
		// cerr << x << endl;
		res++;
	}
}
 
void solve()
{
	cin >> n;
	rep(i, 1, n - 1)
	{
		int u, v;
		cin >> u >> v;
		g[u].push_back(v);
		g[v].push_back(u);
	}
	dfs(1, 0);
	rep(i, 1, n)
	{
		if (i * i > n)
			break;
		if (n % i == 0)
		{
			calc(i);
			if (i * i != n)
				calc(n / i);
		}
	}
	cout << res << endl;
}
 
signed main()
{
	freopen("eat.in", "r", stdin);
	freopen("eat.out", "w", stdout);
	// cerr<<(&en-&st)/1024.0/1024.0<<endl;
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int testcase = 1;
	// cin >> testcase;
	while (testcase--)
		solve();
	return 0;
}

C. 弹弹床

想不到状态的可爱 $dp$ 题捏~~~

对于 $[1,i]$ 这个区间, 必然被从右侧进入再出来一定次数，所以令 $dp_{i,j}$ 表示区间 $[1,i]$ 在若干步以后, 剩下 $j$ 个向右的没有确定终点

d p_{i, j} = {\begin{cases} d p_{i - 1, j} \times j + d p_{i - 1, j - 1}, S_{i} = R \\ d p_{i - 1, j} \times j + d p_{i - 1, j + 1} \times (j + 1) \times j, S_{i} = L \end{cases}

$dp_{i,j}=\begin{cases} dp_{i-1,j} \times j + dp_{i-1,j-1}, S_i=R\\ dp_{i-1,j}\times j +dp_{i-1,j+1} \times (j+1) \times j, S_i=L \end{cases}$

可以获得60分的好成绩

考虑怎么处理中间的数值，发现上述dp可以反过来再做一次, 算出区间 $[i,n]$ , 剩下 $j$ 个向左的方案数

然后枚举终点，左右匹配即可

可爱的code捏

 #include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define double long double
#define mms(arr, n) memset(arr, n, sizeof(arr))
#define rep(i, a, n) for (int i = (a); i <= (n); ++i)
#define per(i, n, a) for (int i = (n); i >= (a); --i)
int max(int a, int b) {
    if (a > b)
        return a;
    return b;
}
int min(int a, int b) {
    if (a < b)
        return a;
    return b;
}
const int INF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1000000007;
const int N = 5e3 + 10;
 
// bool st;
int n;
char s[N];
int f[N][N], g[N][N];
int fac[N];
// bool en;
 
void solve() {
    cin >> n;
    cin >> (s + 1);
    if (n == 1) {
        cout << "1" << endl;
        return;
    }
    f[0][0] = 1;
    rep(i, 1, n) {
        rep(j, 1, i) {
            if (s[i] == 'L')
                f[i][j] = (f[i - 1][j] * j % MOD + f[i - 1][j + 1] * (j + 1) % MOD * j % MOD) % MOD;
            else
                f[i][j] = (f[i - 1][j - 1] + f[i - 1][j] * j % MOD) % MOD;
            f[i][j] %= MOD;
        }
    }
    g[n + 1][0] = 1;
    per(i, n, 1) {
        rep(j, 1, n - i + 1) {
            if (s[i] == 'R')
                g[i][j] = (g[i + 1][j] * j % MOD + g[i + 1][j + 1] * (j + 1) % MOD * j % MOD) % MOD;
            else
                g[i][j] = (g[i + 1][j - 1] + g[i + 1][j] * j % MOD) % MOD;
            g[i][j] %= MOD;
        }
    }
    fac[0] = 1;
    rep(i, 1, n) fac[i] = fac[i - 1] * i % MOD;
    rep(i, 1, n) {
        int res = 0;
        rep(j, 0, i - 1) {
            if (j)
                (res += fac[j] * fac[j - 1] % MOD * f[i - 1][j] % MOD * g[i + 1][j - 1] % MOD) %= MOD;
            (res += fac[j] * fac[j] % MOD * f[i - 1][j] % MOD * g[i + 1][j] % MOD * 2 % MOD) %= MOD;
            (res += fac[j + 1] * fac[j] % MOD * f[i - 1][j] % MOD * g[i + 1][j + 1] % MOD) %= MOD;
        }
        cout << res << ' ';
    }
}
 
signed main() {
    freopen("jump.in", "r", stdin);
    freopen("jump.out", "w", stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int testcase = 1;
    // cin >> testcase;
    while (testcase--) solve();
    return 0;
}

D. 数星星

小丑了上次补这题是ACV的

这道题去年做过我记得 x–r--z- 去年还补过可惜没有什么用 --zc

将查询按照右端点排序，可以考虑把所有贡献算在当前节点的最后一次出现上

那么，当处理到当前右端点时，直接查询左端点即可

然后用一个set去维护每个区间的覆盖情况，用树状数组维护每个时刻的值，树剖做链上修改

巨大的code捏

 #include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define double long double
#define mms(arr, n) memset(arr, n, sizeof(arr))
#define rep(i, a, n) for (int i = (a); i <= (n); ++i)
#define per(i, n, a) for (int i = (n); i >= (a); --i)
#define endl '\n'
int max(int a, int b) {
    if (a > b)
        return a;
    return b;
}
int min(int a, int b) {
    if (a < b)
        return a;
    return b;
}
const int INF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1000000007;
const int N = 1e5 + 10;
 
// bool st;
int n, m, q;
int a[N];
vector<int> g[N];
int dep[N];
int siz[N];
int fa[N];
int top[N];
int son[N], sum[N];
int dfn[N], rnk[N], tot;
// bool en;
 
void dfs(int x, int y) {
    dep[x] = dep[y] + 1;
    siz[x] = 1;
    fa[x] = y;
    for (auto v : g[x]) {
        if (v == y)
            continue;
        dfs(v, x);
        siz[x] += siz[v];
        if (siz[v] > siz[son[x]])
            son[x] = v;
    }
}
 
void dfs2(int x, int tp) {
    dfn[x] = ++tot;
    rnk[tot] = x;
    top[x] = tp;
    sum[tot] = sum[tot - 1] + a[x];
    if (son[x])
        dfs2(son[x], tp);
    for (auto v : g[x]) {
        if (v != son[x] && v != fa[x]) {
            dfs2(v, v);
        }
    }
}
 
struct BIT {
    int tr[N];
 
    void add(int x, int v) {
        while (x) {
            tr[x] += v;
            x -= x & -x;
        }
    }
 
    int sum(int x) {
        int res = 0;
        while (x <= m) {
            res += tr[x];
            x += x & -x;
        }
        return res;
    }
} tr;
 
pii w[N];
 
struct query {
    int l, r, id;
    bool operator<(const query &y) const { return l < y.l; }
} s[N];
 
set<query> st;
 
bool cmp(query a, query b) { return a.r < b.r; }
 
int ans[N];
 
void add(int l, int r, int id) {
    while (1) {
        auto it = st.upper_bound({ r, 0, 0 });
        if (it == st.begin())
            break;
        it--;
        if ((*it).r < l)
            break;
        auto [xl, xr, xid] = (*it);
        // cerr << xl << ' ' << xr << ' ' << xid << endl;
        st.erase(it);
        tr.add(xid, sum[max(l, xl) - 1] - sum[min(r, xr)]);
        if (xl < l)
            st.insert({ xl, l - 1, xid });
        if (xr > r)
            st.insert({ r + 1, xr, xid });
    }
    tr.add(id, sum[r] - sum[l - 1]);
    st.insert({ l, r, id });
}
 
void calc(int u, int v, int id) {
    // cerr << u << ' ' << v << endl;
    while (top[u] != top[v]) {
        if (dfn[top[u]] < dfn[top[v]])
            swap(u, v);
        add(dfn[top[u]], dfn[u], id);
        u = fa[top[u]];
    }
    if (dfn[u] > dfn[v])
        swap(u, v);
    add(dfn[u], dfn[v], id);
}
 
void solve() {
    cin >> n >> m >> q;
    rep(i, 1, n) { cin >> a[i]; }
    rep(i, 1, n - 1) {
        int u, v;
        cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    dfs(1, 0);
    dfs2(1, 1);
    rep(i, 1, m) { cin >> w[i].first >> w[i].second; }
    rep(i, 1, q) {
        cin >> s[i].l >> s[i].r;
        s[i].id = i;
    }
    sort(s + 1, s + q + 1, cmp);
    rep(i, 1, q) {
        rep(j, s[i - 1].r + 1, s[i].r) calc(w[j].first, w[j].second, j);
        ans[s[i].id] = tr.sum(s[i].l);
    }
    rep(i, 1, q) cout << ans[i] << endl;
}
 
signed main() {
    freopen("star.in", "r", stdin);
    freopen("star.out", "w", stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int testcase = 1;
    // cin >> testcase;
    while (testcase--) solve();
    return 0;
}

20231016比赛总结

拿了大众分没有严重的挂分

$\mathcal{O(n\sqrt{n}log^2n)=O(n^2)}$

写代码之前先好好算时间复杂度qwq

2023.10.17

A. 子段排序

构造题注意到排序操作 = $a_i<a_j$ 时的 swap

可爱的code捏

 #include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define double long double
#define mms(arr, n) memset(arr, n, sizeof(arr))
#define rep(i, a, n) for (int i = (a); i <= (n); ++i)
#define per(i, n, a) for (int i = (n); i >= (a); --i)
int max(int a, int b)
{
	if (a > b)
		return a;
	return b;
}
int min(int a, int b)
{
	if (a < b)
		return a;
	return b;
}
const int INF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1000000007;
const int N = 1e3 + 10;
 
// bool st;
int n;
int a[N], b[N];
// bool en;
 
void solve()
{
	cin >> n;
	rep(i, 1, n)
	{
		cin >> a[i];
	}
	rep(i, 1, n) cin >> b[i];
	vector<pii> res;
	rep(i, 1, n)
	{
		if (a[i] != b[i])
		{
			int p = i, mi = INF;
			while (a[p] != b[i] && p <= n)
			{
				mi = min(mi, a[p]);
				p++;
			}
			// cerr << i << ' ' << p << endl;
			if (a[p] > mi || p > n)
			{
				cout << "-1" << endl;
				return;
			}
			per(j, p - 1, i)
			{
				res.push_back({j, j + 1});
				swap(a[j], a[j + 1]);
			}
		}
		// rep(j, 1, n) cerr << a[j] << ' ';
		// cerr << endl;
	}
	cout << "0" << endl;
	cout << res.size() << endl;
	for (auto v : res)
		cout << v.first << ' ' << v.second << endl;
}
 
signed main()
{
	freopen("subarray.in", "r", stdin);
	freopen("subarray.out", "w", stdout);
	// cerr<<(&en-&st)/1024.0/1024.0<<endl;
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int testcase = 1;
	cin >> testcase;
	while (testcase--)
		solve();
	return 0;
}

B. 伪快速排序

$dp_{i,j}$ 表示 $[1,i]$ , 最后一个数为 $j$ 的方案数

d p_{i, j} = \sum_{k = 0}^{i - 1} d p_{k, j - 1} \cdot d p_{i - k - 1, j - 1} \cdot (\binom{i - 1}{k})

$dp_{i,j}=\sum\limits_{k=0}^{i-1} dp_{k,j-1} \cdot dp_{i-k-1,j-1} \cdot \binom{i-1}{k}$

可爱的code捏

 #include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define double long double
#define mms(arr, n) memset(arr, n, sizeof(arr))
#define rep(i, a, n) for (int i = (a); i <= (n); ++i)
#define per(i, n, a) for (int i = (n); i >= (a); --i)
int max(int a, int b)
{
	if (a > b)
		return a;
	return b;
}
int min(int a, int b)
{
	if (a < b)
		return a;
	return b;
}
const int INF = 0x3f3f3f3f3f3f3f3f;
int MOD = 1000000007;
const int N = 3e2 + 10;
 
// bool st;
int dp[N][N];
int c[N][N];
// bool en;
 
void solve()
{
	int n, k;
	cin >> n >> k;
	k = min(k, n);
	cout << dp[n][k] << endl;
}
 
signed main()
{
	freopen("qsort.in","r",stdin);
	freopen("qsort.out","w",stdout);
	// cerr<<(&en-&st)/1024.0/1024.0<<endl;
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int testcase = 1;
	cin >> testcase >> MOD;
	c[0][0] = 1;
	rep(i, 1, 300)
	{
		c[i][0] = c[i][i] = 1;
		rep(j, 1, 300)
		{
			c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % MOD;
		}
		dp[0][i] = 1;
	}
	rep(i, 1, 300)
	{
		dp[i][1] = 1;
		rep(j, 2, 300)
		{
			rep(p, 0, i - 1)
			{
				(dp[i][j] += dp[p][j - 1] * dp[i - p - 1][j - 1] % MOD * c[i - 1][p] % MOD) %= MOD;
			}
		}
	}
	while (testcase--)
		solve();
	return 0;
}

C. 二进制式

表达式题, 先见表达式树, 考虑 dfs 算每个节点为 $1$ 的概率

然后从根向下 dfs 即可算每个点的答案

可爱的code捏

 #include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define double long double
#define mms(arr, n) memset(arr, n, sizeof(arr))
#define rep(i, a, n) for (int i = (a); i <= (n); ++i)
#define per(i, n, a) for (int i = (n); i >= (a); --i)
int max(int a, int b)
{
	if (a > b)
		return a;
	return b;
}
int min(int a, int b)
{
	if (a < b)
		return a;
	return b;
}
const int INF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 998244353;
const int N = 2e5 + 10;
 
// bool st;
int n, len;
char s[N << 4];
struct node
{
	int l, r;
	int op, val; // 1 and 2 or 3 xor
} p[N << 2];
int tot = 0, cnt = 0;
stack<int> st, ope;
int res[N];
// bool en;
 
int qpow(int a, int b)
{
	int res = 1;
	while (b)
	{
		if (b & 1)
			res = res * a % MOD;
		a = a * a % MOD;
		b >>= 1;
	}
	return res;
}
 
void dfs(int x)
{
	if (p[x].l > n)
		dfs(p[x].l);
	if (p[x].r > n)
		dfs(p[x].r);
	if (p[x].op == 1)
		p[x].val = p[p[x].l].val * p[p[x].r].val % MOD;
	else if (p[x].op == 2)
		p[x].val = (1 - (1 - p[p[x].l].val + MOD) % MOD * (1 - p[p[x].r].val + MOD) % MOD + MOD) % MOD;
	else
		p[x].val = ((1 - p[p[x].l].val + MOD) % MOD * p[p[x].r].val % MOD + (1 - p[p[x].r].val + MOD) % MOD * p[p[x].l].val % MOD) % MOD;
}
 
void calc(int x, int v)
{
	if (x <= n)
	{
		res[x] = v;
		return;
	}
	if (p[x].op == 1)
	{
		calc(p[x].l, v * p[p[x].r].val % MOD);
		calc(p[x].r, v * p[p[x].l].val % MOD);
	}
	else if (p[x].op == 2)
	{
		calc(p[x].l, v * (1 - p[p[x].r].val + MOD) % MOD);
		calc(p[x].r, v * (1 - p[p[x].l].val + MOD) % MOD);
	}
	else
	{
		calc(p[x].l, v);
		calc(p[x].r, v);
	}
}
 
void solve()
{
	cin >> n >> (s + 1);
	len = strlen(s + 1);
	cnt = n;
	int inv2 = qpow(2, MOD - 2);
	rep(i, 1, n)
	{
		p[i].val = inv2;
	}
	rep(i, 1, len)
	{
		if (s[i] == 'x')
			st.push(++tot);
		else if (s[i] == '&')
			ope.push(1);
		else if (s[i] == '|')
			ope.push(2);
		else if (s[i] == '^')
			ope.push(3);
		else if (s[i] == ')')
		{
			int b = st.top();
			st.pop();
			int a = st.top();
			st.pop();
			int opr = ope.top();
			ope.pop();
			p[++cnt] = {a, b, opr};
			st.push(cnt);
		}
	}
	dfs(cnt);
	calc(cnt, 1);
	rep(i, 1, n)
	{
		cout << res[i] << endl;
	}
}
 
signed main()
{
	freopen("binary.in","r",stdin);
	freopen("binary.out","w",stdout);
	// cerr<<(&en-&st)/1024.0/1024.0<<endl;
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int testcase = 1;
	// cin >> testcase;
	while (testcase--)
		solve();
	return 0;
}

D. 网格图

挖坑待填

20231017比赛总结

除了最后一题没有挂分

认真看数据范围，下面不一定包括上面！！！

2023.10.18

A. 玛雅历

形似儒略日, 稍微简单一点, 注意到公元前很烦, 考虑直接将公元前向后平移 $1$ 年, 然后使公元 $0$ 年存在

可爱的code捏

 #include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define double long double
#define mms(arr, n) memset(arr, n, sizeof(arr))
#define rep(i, a, n) for (int i = (a); i <= (n); ++i)
#define per(i, n, a) for (int i = (n); i >= (a); --i)
int max(int a, int b)
{
	if (a > b)
		return a;
	return b;
}
int min(int a, int b)
{
	if (a < b)
		return a;
	return b;
}
const int INF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1000000007;
const int N = 2e5 + 10;
 
// bool st;
int cnt400 = 0;
int a[N];
int num[10] = {0, 23040000000, 1152000000, 57600000, 2880000, 144000, 7200, 360, 20, 1};
int month[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
// bool en;
 
void solve()
{
	rep(i, 1, 8)
	{
		cin >> a[i];
		char c;
		cin >> c;
	}
	cin >> a[9];
	int daycnt = 0;
	rep(i, 1, 9) daycnt = daycnt + num[i] * a[i];
	// cerr << daycnt << endl;
	int yearcnt = daycnt / cnt400;
	daycnt %= cnt400;
	int yearnum = -3113 + yearcnt * 400, monthnum = 8;
	int curyearnum = 0;
	if ((yearnum + 1) % 4 == 0 && ((yearnum + 1) % 100 != 0 || (yearnum + 1) % 400 == 0))
		curyearnum = 366;
	else
		curyearnum = 365;
	while (daycnt >= curyearnum)
	{
		daycnt -= curyearnum;
		yearnum++;
		if ((yearnum + 1) % 4 == 0 && ((yearnum + 1) % 100 != 0 || (yearnum + 1) % 400 == 0))
			curyearnum = 366;
		else
			curyearnum = 365;
	}
	if (yearnum % 4 == 0 && (yearnum % 100 != 0 || yearnum % 400 == 0))
		month[2] = 29;
	while (daycnt >= month[monthnum])
	{
		daycnt -= month[monthnum];
		monthnum++;
		if (monthnum > 12)
		{
			monthnum = 1;
			yearnum++;
			if (yearnum % 4 == 0 && (yearnum % 100 != 0 || yearnum % 400 == 0))
				month[2] = 29;
			else
				month[2] = 28;
		}
	}
	int daynum = 11;
	while (daycnt)
	{
		daycnt--;
		daynum++;
		if (daynum > month[monthnum])
		{
			daynum = 1;
			monthnum++;
			if (monthnum > 12)
			{
				monthnum = 1;
				yearnum++;
				if (yearnum % 4 == 0 && (yearnum % 100 != 0 || yearnum % 400 == 0))
					month[2] = 29;
				else
					month[2] = 28;
			}
		}
	}
	if (yearnum <= 0)
		cout << daynum << ' ' << monthnum << ' ' << -(yearnum - 1) << " BC" << endl;
	else
		cout << daynum << ' ' << monthnum << ' ' << yearnum << endl;
}
 
signed main()
{
	freopen("maya.in", "r", stdin);
	freopen("maya.out", "w", stdout);
	// cerr<<(&en-&st)/1024.0/1024.0<<endl;
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int testcase = 1;
	cin >> testcase;
	rep(i, 0, 399)
	{
		if (i % 4 == 0 && (i % 100 != 0 || i % 400 == 0))
			cnt400 += 366;
		else
			cnt400 += 365;
	}
	// cerr << cnt400 << endl;
	while (testcase--)
		solve();
	return 0;
}

B. 大融合

如题, 大融合, 可以拆成 $2$ 题

第一部分算最小的 $S$ , 直接map即可, 考场差点写 trie

第二部分显然二分, 考虑check, 可以令 $dp_{i,j}$ 表示到第 $i$ 行, 第 $j$ 列最多获得多少

显然这个 $dp$ 可以单调队列优化

发现 $3$ 个机器人分别走分别贡献其实是诈骗条件, 和只有一个走并贡献 $100\%$ 等价

可爱的code捏

 #include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define double long double
#define mms(arr, n) memset(arr, n, sizeof(arr))
#define rep(i, a, n) for (int i = (a); i <= (n); ++i)
#define per(i, n, a) for (int i = (n); i >= (a); --i)
int max(int a, int b)
{
	if (a > b)
		return a;
	return b;
}
int min(int a, int b)
{
	if (a < b)
		return a;
	return b;
}
const int INF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1000000007;
const int N = 1e3 + 10;
 
// bool st;
int n, m, q, d;
map<int, bool> mp;
int minS;
int dp[N][N];
int a[N][N], col[N][N];
// bool en;
 
int solveminS()
{
	cin >> n >> m >> q >> d;
	rep(i, 1, m)
	{
		int x;
		cin >> x;
		int tmp = 10;
		while (tmp <= x * 10)
		{
			// cerr << x % tmp << endl;
			mp[x % tmp] = true;
			tmp *= 10;
		}
	}
	int res = 0;
	rep(i, 1, q)
	{
		int l, b, v;
		cin >> l >> b >> v;
		if (!mp.count(b))
			res += v;
	}
	return res;
}
 
bool check(int x)
{
	int ld = max(1, d - x), rd = min(n, d + x);
	memset(dp, -0x3f, sizeof(dp));
	dp[1][1] = 0;
	deque<pii> que[N][2];
	int res = 0;
	rep(i, 1, n)
	{
		deque<pii> cur[2];
		rep(j, 1, n)
		{
			while (!cur[0].empty() && cur[0].front().second < j - rd)
				cur[0].pop_front();
			while (!cur[1].empty() && cur[1].front().second < j - rd)
				cur[1].pop_front();
			while (!que[j][0].empty() && que[j][0].front().second < j - rd)
				que[j][0].pop_front();
			while (!que[j][1].empty() && que[j][1].front().second < j - rd)
				que[j][1].pop_front();
			if (j > ld)
			{
				while (!cur[col[i][j - ld]].empty() && cur[col[i][j - ld]].back().first <= dp[i][j - ld])
					cur[col[i][j - ld]].pop_back();
				cur[col[i][j - ld]].push_back({dp[i][j - ld], j - ld});
			}
			if (i > ld)
			{
				while (!que[j][col[i - ld][j]].empty() && que[j][col[i - ld][j]].back().first <= dp[i - ld][j])
					que[j][col[i - ld][j]].pop_back();
				que[j][col[i - ld][j]].push_back({dp[i - ld][j], i - ld});
			}
			if (!cur[0].empty())
				dp[i][j] = max(dp[i][j], cur[0].front().first + a[i][j] + (col[i][j] == 0));
			if (!cur[1].empty())
				dp[i][j] = max(dp[i][j], cur[1].front().first + a[i][j] + (col[i][j] == 1));
			if (!que[j][0].empty())
				dp[i][j] = max(dp[i][j], que[j][0].front().first + a[i][j] + (col[i][j] == 0));
			if (!que[j][1].empty())
				dp[i][j] = max(dp[i][j], que[j][1].front().first + a[i][j] + (col[i][j] == 1));
			res = max(res, dp[i][j]);
			// cerr << dp[i][j] << ' ';
		}
		// cerr << endl;
	}
	// cerr << x << ' ' << res << endl;
	return res >= minS;
}
 
void solve()
{
	rep(i, 1, n)
	{
		rep(j, 1, n)
		{
			cin >> col[i][j];
			col[i][j]--;
		}
	}
	rep(i, 1, n)
	{
		rep(j, 1, n)
		{
			cin >> a[i][j];
		}
	}
	int l = 0, r = n;
	while (l < r)
	{
		int mid = l + r >> 1;
		if (check(mid))
			r = mid;
		else
			l = mid + 1;
	}
	if (check(r))
		cout << r << endl;
	else
		cout << "-1" << endl;
}
 
signed main()
{
	freopen("harbinbeer.in", "r", stdin);
	freopen("harbinbeer.out", "w", stdout);
	// cerr<<(&en-&st)/1024.0/1024.0<<endl;
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int testcase = 1;
	// cin >> testcase;
	minS = solveminS();
	// cerr << minS << endl;
	solve();
	return 0;
}

C. 好 ♂ 朋 ♂ 友

卡常狗能不能死一死啊

很巧妙的性质: 同一个左端点的所有区间的 or 值最多只有 $30$ 种情况

对于每个左端点, 二分或者倍增+st表, 算出每一段的位置和值, 对于合法的段区间 $+1$

将查询离线, 按左端点排序, 对于每一个查询, 先将 $ql$ 左侧的节点作为左端点的贡献都去掉, 再在线段树上查询当前的答案

被卡常了qwq

被卡常的巨大code捏

 #include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define double long double
#define mms(arr, n) memset(arr, n, sizeof(arr))
#define rep(i, a, n) for (int i = (a); i <= (n); ++i)
#define per(i, n, a) for (int i = (n); i >= (a); --i)
int max(int a, int b)
{
	if (a > b)
		return a;
	return b;
}
int min(int a, int b)
{
	if (a < b)
		return a;
	return b;
}
const int INF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1000000007;
const int N = 1e5 + 10;
const int M = 1e6 + 10;
 
// bool st;
int n, k, t;
int a[N];
set<int> s;
int res[M];
int st[N][20];
int lg[N];
pair<pii, int> q[M];
vector<pii> vec[N];
// bool en;
 
struct segment_tree
{
#define ls p << 1
#define rs p << 1 | 1
	struct node
	{
		int sum, laz;
	} tr[N << 2];
 
	void pushup(int p)
	{
		tr[p].sum = tr[ls].sum + tr[rs].sum;
	}
 
	void pushdown(int p, int l, int r)
	{
		int mid = l + r >> 1;
		tr[ls].sum += (mid - l + 1) * tr[p].laz;
		tr[ls].laz += tr[p].laz;
		tr[rs].sum += (r - mid) * tr[p].laz;
		tr[rs].laz += tr[p].laz;
		tr[p].laz = 0;
	}
 
	void update(int p, int l, int r, int ll, int rr, int val)
	{
		if (ll <= l && r <= rr)
		{
			tr[p].sum += (r - l + 1) * val;
			tr[p].laz += val;
			return;
		}
		pushdown(p, l, r);
		int mid = l + r >> 1;
		if (ll <= mid)
			update(ls, l, mid, ll, rr, val);
		if (rr > mid)
			update(rs, mid + 1, r, ll, rr, val);
		pushup(p);
	}
 
	int query(int p, int l, int r, int ll, int rr)
	{
		if (ll <= l && r <= rr)
			return tr[p].sum;
		pushdown(p, l, r);
		int mid = l + r >> 1;
		int res = 0;
		if (ll <= mid)
			res += query(ls, l, mid, ll, rr);
		if (rr > mid)
			res += query(rs, mid + 1, r, ll, rr);
		return res;
	}
} seg;
 
int get(int l, int r)
{
	int t = lg[r - l + 1];
	return (st[l][t] | st[r - (1 << t) + 1][t]);
}
 
void solve()
{
	cin >> n >> t >> k;
	rep(i, 1, k)
	{
		int x;
		cin >> x;
		s.insert(x);
	}
	lg[1] = 0;
	rep(i, 1, n)
	{
		if (i != 1)
			lg[i] = lg[i >> 1] + 1;
		cin >> a[i];
		st[i][0] = a[i];
	}
	rep(i, 1, 19)
	{
		for (int j = 1; j + (1 << i) - 1 <= n; j++)
		{
			st[j][i] = st[j][i - 1] | st[j + (1 << i - 1)][i - 1];
		}
	}
	rep(i, 1, n)
	{
		int la = i - 1;
		int j = i;
		while (j <= n)
		{
			per(p, 19, 0)
			{
				if (j + (1 << p) > n)
					continue;
				if (get(i, j + (1 << p)) == get(i, j))
					j = j + (1 << p);
			}
			if (s.find(get(i, j) % 10) != s.end())
			{
				vec[i].push_back({la + 1, j});
				seg.update(1, 1, n, la + 1, j, 1);
			}
			// cerr << i << ' ' << la + 1 << ' ' << j << endl;
			la = j;
			j++;
		}
	}
	rep(i, 1, t)
	{
		cin >> q[i].first.first >> q[i].first.second;
		q[i].second = i;
	}
	sort(q + 1, q + t + 1);
	int p = 1;
	rep(i, 1, t)
	{
		// cerr << q[i].first.first << ' ' << q[i].first.second << endl;
		while (p < q[i].first.first)
		{
			for (auto v : vec[p])
				seg.update(1, 1, n, v.first, v.second, -1);
			p++;
		}
		res[q[i].second] = seg.query(1, 1, n, q[i].first.first, q[i].first.second);
	}
	rep(i, 1, t) cout << res[i] << '\n';
}
 
signed main()
{
	freopen("friend.in", "r", stdin);
	freopen("friend.out", "w", stdout);
	// cerr<<(&en-&st)/1024.0/1024.0<<endl;
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int testcase = 1;
	// cin >> testcase;
	while (testcase--)
		solve();
	return 0;
}

D. 图的直径

挖坑待补

20231018比赛总结

前 $2$ 题还好, T1第一遍写忽略了很多细节, T2假的代码过了乐

T3注意常数(卡常狗死一死)

能拿的分应该都拿到了吧

2023.10.19

A. 潜艇

上一篇CSP2023感触回忆录

本文作者：xiaruize's Blog

本文链接：https://www.cnblogs.com/xiaruize/p/17768600.html

posted @ 2023-10-16 23:00 xiaruize 阅读(54) 评论(3) 编辑收藏举报

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xiaruize's Blog

CSP模拟赛记录

CSP模拟赛记录

2023.10.15

A. 超速

B. 对常规的斗争

C. 海报

D. 机器人锦标赛

2023.10.15比赛总结

2023.10.16

A. 魔力子串

B. 吃树

C. 弹弹床

D. 数星星

20231016比赛总结

2023.10.17

A. 子段排序

B. 伪快速排序

C. 二进制式

D. 网格图

20231017比赛总结

2023.10.18

A. 玛雅历

B. 大融合

C. 好 ♂ 朋 ♂ 友

D. 图的直径

20231018比赛总结

2023.10.19

A. 潜艇

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	// Author: xiaruize
	#include <bits/stdc++.h>
	using namespace std;
	#define int long long
	#define ull unsigned long long
	#define ALL(a) (a).begin(), (a).end()
	#define pb push_back
	#define mk make_pair
	#define pii pair<int, int>
	#define pis pair<int, string>
	#define sec second
	#define fir first
	#define sz(a) int((a).size())
	#define Yes cout << "Yes" << endl
	#define YES cout << "YES" << endl
	#define No cout << "No" << endl
	#define NO cout << "NO" << endl
	#define debug(x) cerr << #x << ": " << x << endl
	#define mms(arr, n) memset(arr, n, sizeof(arr))
	#define rep(i, a, n) for (int i = (a); i <= (n); ++i)
	#define per(i, n, a) for (int i = (n); i >= (a); --i)
	int max(int a, int b) {
	if (a > b)
	return a;
	return b;
	}
	int min(int a, int b) {
	if (a < b)
	return a;
	return b;
	}
	void chkmx(int &x, int y) {
	if (y > x)
	x = y;
	}
	void chkmi(int &x, int y) {
	if (y < x)
	x = y;
	}
	const int INF = 0x3f3f3f3f3f3f3f3f;
	const int MOD = 1000000007;
	const int N = 2e5 + 10;

	// bool st;
	int n;
	int v[N], l[N];
	int m;
	int a[N], f[N];
	int q;
	int lim;
	// bool en;

	bool check(int x) {
	double t = 0;
	rep(i, 1, n) { t = t + (double)l[i] / (v[i] + a[x]); }
	return t < lim;
	}

	void solve() {
	int s, t;
	cin >> s >> t;
	lim = t - s;
	int l = 0, r = m;
	while (l < r) {
	int mid = l + r >> 1;
	if (check(mid))
	r = mid;
	else
	l = mid + 1;
	}
	cout << f[r] << endl;
	}

	signed main() {
	freopen("speed.in", "r", stdin);
	freopen("speed.out", "w", stdout);
	// cerr<<(&en-&st)/1024.0/1024.0<<endl;
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int testcase = 1;
	cin >> n;
	rep(i, 1, n) cin >> v[i];
	rep(i, 1, n) cin >> l[i];
	cin >> m;
	a[m + 1] = 1e9;
	rep(i, 1, m - 1) cin >> a[i];
	rep(i, 1, m) cin >> f[i];
	cin >> testcase;
	while (testcase--) solve();
	return 0;
	}