C. Constructing Ranches

给一棵树，点带权，问有多少条路径的点权能构成一个严格(指有面积)的多边形。

结论：充要条件为∑ai−max⁡ai>max⁡ai\sum a_i - \max{a_i} > \max{a_i}∑ai−maxai>maxai

考虑点分治, 处理出当前分治子树下所有点到分治点的路径中的权值和以及权值最大值.

然后按权值最大值排序，用树状数组计数，复杂度为O(nlog2n)O(nlog^2n)O(nlog2n).

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, ll> pil;
#define fi first
#define se second
#define all(x) (x).begin(),(x).end()
const int N = 200010;vector<int> E[N];
vector<pil> info;
int sz[N], a[N];
bool vis[N];void prepare(int u, int fa) {sz[u] = 1;for(auto &v : E[u]) {if(v == fa || vis[v]) continue;prepare(v, u);sz[u] += sz[v];}
}
void getroot(int u, int fa, int &m, int &root) {if(sz[u] * 2 < m) return;if(sz[u] < sz[root]) root = u;for(auto &v : E[u]) {if(v == fa || vis[v]) continue;getroot(v, u, m, root);}
}
void getinfo(vector<pil> &info, int u, int fa, int nowmx, ll nowsum) {nowmx = max(nowmx, a[u]), nowsum = nowsum + a[u];for(auto &v : E[u]) {if(v == fa || vis[v]) continue;getinfo(info, v, u, nowmx, nowsum);}info.emplace_back(nowmx, nowsum);
}
ll calc(int u, int root) {info.clear();int _mx = u == root ? 0 : a[root], _a = _mx;getinfo(info, u, 0, _mx, _a);
//  info.pop_back();vector<ll> dec(info.size());for(int i = 0; i < (int)info.size(); i++) {dec[i] = info[i].se - a[root];}sort(all(dec)); dec.erase(unique(all(dec)), dec.end());vector<int> bit(dec.size() + 1, 0);int L = dec.size();auto add = [&](int x) { for(; x <= L; x += x&-x) bit[x]++;};auto sum = [&](int x) { int ret = 0; for(; x; x -= x&-x) ret += bit[x]; return ret;};sort(all(info), greater<pil>());ll ret = 0;for(auto &e : info) {ret += sum(upper_bound(all(dec),e.se-a[root])-dec.begin());int idx = upper_bound(all(dec),2*e.fi-e.se)-dec.begin()+1;if(idx <= L) add(idx);}return ret;
}
ll divide(int u) {prepare(u, 0);int m = sz[u], root = u;getroot(u, 0, m, root);ll ret = calc(root, root); vis[root] = 1;
//    cout << root << ' ' << ret << '\n';
//    for(auto &e : info) cout << e.fi << ' ' << e.se << endl;for(auto &v : E[root]) {if(vis[v]) continue;ret -= calc(v, root);ret += divide(v);}return ret;
}void solve(int cas) {int n; cin >> n;for(int i = 1; i <= n; i++) E[i].clear(), vis[i] = 0;for(int i = 1; i <= n; i++) cin >> a[i];for(int i = 1, u, v; i < n; i++) {cin >> u >> v;E[u].emplace_back(v);E[v].emplace_back(u);}cout << divide(1) << '\n';
}int main() {#ifdef localfreopen("in.txt", "r", stdin);
#endifios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int _; cin >> _;for(int i = 1; i <= _; i++) solve(i);return 0;
}

E. Erasing Numbers

给一个111到nnn的排列，保证nnn一定是奇数.

每次你可以选择连续的三个数，并将三个数中的最大值和最小值删掉，删到只有最后一个数为止，

问第iii个数能否留在最后.

考虑比第iii个数大的数的个数和小的数的个数，如果能删到让两种数的个数相等，则第iii个数可以留到最后。

贪心删个数多的一边即可，复杂度O(n2)O(n^2)O(n2)。

#include<bits/stdc++.h>
using namespace std;
typedef pair<int, int> P;
#define F first
#define S second
#define all(x) (x).begin(),(x).end()
const int N = 5010;
int a[N], n;int go(int l, int r, int p) {if(l > r) return 0;int num1 = a[p] - 1, num2 = n - a[p];int ret = 0, sz = 0;if(num1 < num2) {for(int i = l; i <= r; i++) {if(a[i] > a[p]) {if(++sz == 3) sz -= 2, ret++;} else if(--sz == -1) sz = 0;}} else {for(int i = l; i <= r; i++) {if(a[i] < a[p]) {if(++sz == 3) sz -= 2, ret++;} else if(--sz == -1) sz = 0;}}return ret;
}void solve(int cas) {cin >> n;for(int i = 0; i < n; i++) cin >> a[i];for(int i = 0; i < n; i++) {int c1 = go(0, i - 1, i);int c2 = go(i + 1, n - 1, i);int c3 = abs(n + 1 - 2 * a[i]) / 2;if(c1 + c2 >= c3) cout << "1";else cout << "0";}cout << '\n';
}int main() {#ifdef localfreopen("in.txt", "r", stdin);
#endifios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int _; cin >> _;for(int i = 0; i < _; i++) solve(i);return 0;
}

H. Hold the Line

两种操作：

1.在xxx位置插入一个yyy;

2.查询[l,r][l,r][l,r]区间中离xxx近的yyy,输出∣x−y∣|x-y|∣x−y∣.

线段树套std::setstd::setstd::set的在线做法常数过大，考虑cdqcdqcdq分治，线段树套单调栈等离线做法.

加快读勉强卡过去.

#include<bits/stdc++.h>
using namespace std;
const int N = 1 << 19 | 7;
const int inf = 0x3f3f3f3f;
struct seg_t {#define ls o << 1
#define rs o<<1|1static int n;static void set_n(int _n) { n = _n;}struct node_t {int val;node_t() {}node_t(int _val):val(_val) {}operator int() { return val;}void init(int _val) { val = _val;}node_t operator + (const node_t &rhs) {return node_t(max(val, rhs.val));}} node[N << 1];void init() {build();}void build(int o = 1, int l = 1, int r = n) {if(!node[o].val) return;node[o].val = 0;if(l == r) return;int mid = (l + r) >> 1;build(ls, l, mid);build(rs, mid + 1, r);// node[o] = node[ls] + node[rs];}void update(int x, int v, int o = 1, int l = 1, int r = n) {if(l == r) return node[o].init(v);int mid = (l + r) >> 1;if(mid >= x) update(x, v, ls, l, mid);else update(x, v, rs, mid + 1, r);node[o] = node[ls] + node[rs];}node_t query(int ql, int qr, int o = 1, int l = 1, int r = n) {if(ql <= l && r <= qr) return node[o];node_t ret = 0; int mid = (l + r) >> 1;if(ql <= mid) ret = ret + query(ql, qr, ls, l, mid);if(qr > mid) ret = ret + query(ql, qr, rs, mid + 1, r);return ret;}
} tree;
int seg_t::n;int ans[N * 2];
struct event {int ty, l, r, h, id;bool operator < (const event&rhs)const {if(h != rhs.h) return h < rhs.h;return ty < rhs.ty;}
} E[N * 2], tE[N * 2];void work(int l, int r) {if(l == r) return;if(r - l < 200) {for(int i = l; i <= r; i++) if(E[i].ty) {for(int j = l; j < i; j++) if(!E[j].ty && E[j].h <= E[i].h) {if(E[i].l <= E[j].l && E[j].l <= E[i].r)ans[E[i].id] = min(ans[E[i].id], E[i].h - E[j].h);}}sort(E + l, E + r + 1);return;}int mid = (l + r) >> 1;work(l, mid); work(mid + 1, r);int p = l, q = mid + 1, c = l;while(q <= r) {if(p <= mid && E[p] < E[q]) {if(!E[p].ty) tree.update(E[p].l, E[p].h);tE[c++] = E[p++];} else {if(E[q].ty) {int t = tree.query(E[q].l, E[q].r);if(t) ans[E[q].id] = min(ans[E[q].id], E[q].h - t);}tE[c++] = E[q++];}}while(p <= mid) tE[c++] = E[p++];for(int i = l; i <= r; i++) E[i] = tE[i];tree.init();
}
inline char nc() {#define SZ 1000000static char buf[SZ], *p1 = buf, *p2 = buf;return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, SZ, stdin), p1 == p2) ? EOF : *p1++;
#undef SZ
}
int scan(int &x) {char c;while(!isdigit(c = nc()) && c != EOF);if(c == EOF) return c;for(x = 0; isdigit(c); c = nc())x = x * 10 + (c ^ 48);return 1;
}
template<typename ...Args>
int scan(int &x, Args&...args) {return scan(x), scan(args...);
}
void print(int x) {static char buf[21], *p2;if(x < 0) putchar('-'), print(-x);else {*(p2 = buf + 20) = '\0';if(!x) *--p2 = 48;while(x) *--p2 = x % 10 ^ 48, x /= 10;puts(p2);}
}
int main() {#ifdef localfreopen("in.txt", "r", stdin);
#endifint _; scan(_);while(_--) {int n, q; scan(n, q);for(int i = 0; i < q; i++) {ans[i] = inf;scan(E[i].ty);if(E[i].ty) scan(E[i].l, E[i].r, E[i].h);else scan(E[i].l, E[i].h), E[i].r = E[i].l;E[i].id = i;}seg_t::set_n(n);work(0, q - 1);for(int i = 0; i < q; i++) tE[E[i].id] = E[i];for(int i = 0; i < q; i++) E[i] = tE[i];for(int i = 0; i < q; i++) E[i].h = 1e9 + 1 - E[i].h;work(0, q - 1);for(int i = 0; i < q; i++) tE[E[i].id] = E[i];for(int i = 0; i < q; i++) E[i] = tE[i];for(int i = 0; i < q; i++) {if(E[i].ty) {if(ans[i] == inf) ans[i] = -1;print(ans[i]);}}}return 0;
}

I. Incoming Asteroids

两种操作:

在q[1..k](1≤k≤3)q[1..k](1\leq k \leq 3)q[1..k](1≤k≤3)个容器里插入一个值为y的元素，当这个值y在一个容器改变时，它在其他容器的值也会改变。
将第xxx个容器的值全部减yyy，若有元素的值小于000，则按插入顺序输出他们，并从容器中删除。

两种操作均强制在线。

据CJB所说，这个idea来自2018年毛营.

将yyy分成kkk份，分别插入到各个容器中，若yyy在某个容器中的值小于000则从其他容器中取出y，然后分成kkk份，重新插入到各个容器中，

每个容器用堆维护最小值，这样每次取出yyy，yyy都会至少减小原来的13\frac1331。均摊复杂度为O(mlog⁡nlog⁡y).O(m\log n\log y).O(mlognlogy).

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define all(x) (x).begin(),(x).end()
const int N = 200010;
int tot = 0;
vector<int> ans;
vector<int> p[N];
vector<ll> g[N];
ll del[N];
int val[N], vis[N];
struct event {ll v;int t, id;event() {}event(ll _v, int _t, int _id) { v = _v, t = _t, id = _id;}bool operator < (const event &r)const {return v > r.v;}
};
priority_queue<event> A[N];
int main() {#ifdef localfreopen("in.txt", "r", stdin);
#endifios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int n, q; cin >> n >> q;while(q--) {int op; cin >> op;if(op == 1) {int id = ++tot;int y, k; cin >> y >> k; y ^= ans.size();val[id] = y;p[id].resize(k);g[id].resize(k);y = (y + k - 1) / k;for(int i = 0; i < k; i++) {int x; cin >> x; x ^= ans.size();A[x].emplace(del[x] + y, vis[id], id);p[id][i] = x;g[id][i] = del[x];}} else {int x, y; cin >> x >> y;x ^= ans.size(), y ^= ans.size();ans.clear();del[x] += y;while(!A[x].empty() && A[x].top().v <= del[x]) {event temp = A[x].top(); A[x].pop();if(temp.t != vis[temp.id]) continue;vis[temp.id]++;ll tt = 0;for(int i = 0; i < (int) p[temp.id].size(); i++) {int x = p[temp.id][i];tt += del[x] - g[temp.id][i];}if(tt >= val[temp.id]) ans.push_back(temp.id);else {val[temp.id] -= tt;int y = (val[temp.id] + p[temp.id].size() - 1) / p[temp.id].size();for(int i = 0; i < (int) p[temp.id].size(); i++) {int x = p[temp.id][i];A[x].emplace(del[x] + y, vis[temp.id], temp.id);g[temp.id][i] = del[x];}}}cout << ans.size();sort(all(ans));for(auto &e : ans) cout << ' ' << e; cout << '\n';}}return 0;
}

J. Junior Mathematician

数位dpdpdp出线题（逃）

#include<bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int N = 5010;
char s1[N], s2[N];
int dp[N][60][60], num[N], pw10[N], n, m;
int dfs(int p, int dt, int prf, bool limit) {if(p == -1) return dt == 0;if(!limit && dp[p][dt][prf] != -1) return dp[p][dt][prf];int up = limit ? num[p] : 9, res = 0;for(int i = 0; i <= up; i++)res=(res+dfs(p-1,(dt+i*pw10[p]-i*prf%m+m)%m,(prf+i)%m,limit&&(i==up)))%mod;if(!limit) dp[p][dt][prf] = res;return res;
}
bool check(char *s) {int n = strlen(s), x = 0, sum = 0, prf = 0;for(int i = 0; i < n; i++) {x = (x * 10 + s[i] - '0') % m;sum = (sum + (s[i] - '0') * prf) % m;prf = (prf + s[i] - '0') % m;}return x == sum;
}
int solve(char *s) {int n = strlen(s);reverse(s, s + n);for(int i = 0; i < n; i++) num[i] = s[i] - '0';return dfs(n - 1, 0, 0, 1);
}int main() {#ifdef localfreopen("in.txt", "r", stdin);
#endifios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int _; cin >> _; pw10[0] = 1;while(_--) {cin >> s1 >> s2 >> m;int n = strlen(s2);memset(dp, -1, sizeof(dp[0]) * n);for(int i = 1; i <= n; i++) {pw10[i] = pw10[i - 1] * 10 % m;}int res = check(s1);res = (res + solve(s2)) % mod;res = (res + mod - solve(s1)) % mod;cout << res << '\n';}return 0;
}

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