Fun is Counting
我们可以发现数组 \(a\) 必须是 \(x\) 或 \(x - 1\),然后分类讨论即可
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1e6 + 5, mod = 998244353;
int inv[N], f[N], g[N], t, n, a[N];
int C(int a, int b) {
if (a < b) {
return 0;
}
return f[a] % mod * g[b] % mod * g[a - b] % mod;
}
void Solve() {
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> a[i];
}
sort(a + 1, a + n + 1);
if (a[1] + 1 < a[n]) {
cout << "0\n";
return ;
}
if (a[1] + 1 == a[n]) {
int cnt = 0;
for (int i = 1; i <= n; i++) {
cnt += (a[i] == a[1]);
}
if (a[n] < cnt + 1) {
cout << "0\n";
return ;
}
cout << C(n, a[n]) * C(a[n], cnt) % mod * C(n - a[n] - 1, (a[n] - cnt) - 1) % mod << "\n";
}
else {
if (a[1] == 1) {
cout << n + (n == 2) << "\n";
return ;
}
if (a[1] == n - 1) {
cout << "1\n";
return ;
}
cout << C(n - a[1] - 1, a[1] - 1) * C(n, a[1]) % mod << "\n";
}
}
signed main() {
ios::sync_with_stdio(0);
cin.tie(0);
inv[1] = f[0] = g[0] = 1;
for (int i = 1; i <= 1000000; i++) {
f[i] = f[i - 1] * i % mod;
if (i > 1) {
inv[i] = inv[mod % i] * (mod - mod / i) % mod;
}
g[i] = g[i - 1] * inv[i] % mod;
}
cin >> t;
while (t--) {
Solve();
}
return 0;
}
Mad MAD Sum II
我们可以画出一个抽象的图,来表示他的值得变化
然后就可以想到一些更抽象的操作
那么我们就可以根据图所示,更改一段区间的值,有线段树维护即可
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 2e5 + 5, INF = 1e18;
struct node {
int x, sum, lazy;
}tr[N * 4];
int t, n, a[N];
map<int, int> mp;
node Merge(node x, node y) {
return {min(x.x, y.x), x.sum + y.sum, INF};
}
void pushdown(int i, int l, int r, int mid) {
if (tr[i].lazy != INF) {
tr[i * 2] = {tr[i].lazy, (tr[i].lazy) * (mid - l + 1), tr[i].lazy};
tr[i * 2 + 1] = {tr[i].lazy, (tr[i].lazy) * (r - mid), tr[i].lazy};
}
tr[i].lazy = INF;
}
void modify(int i, int l, int r, int x, int y, int z) {
if (l > y || r < x) {
return ;
}
if (l >= x && r <= y) {
tr[i] = {z, z * (r - l + 1), z};
return ;
}
int mid = (l + r) >> 1;
pushdown(i, l, r, mid);
modify(i * 2, l, mid, x, y, z);
modify(i * 2 + 1, mid + 1, r, x, y, z);
tr[i] = Merge(tr[i * 2], tr[i * 2 + 1]);
}
void build(int i, int l, int r) {
if (l == r) {
tr[i] = {INF, 0, INF};
return ;
}
int mid = (l + r) >> 1;
build(i * 2, l, mid);
build(i * 2 + 1, mid + 1, r);
tr[i] = {INF, 0, INF};
}
int query(int i, int l, int r, int x, int y) {
if (l > y || r < x) {
return 0;
}
if (l >= x && r <= y) {
return tr[i].sum;
}
int mid = (l + r) >> 1;
pushdown(i, l, r, mid);
return query(i * 2, l, mid, x, y) + query(i * 2 + 1, mid + 1, r, x, y);
}
int Find(int i, int l, int r, int x) {
if (l == r) {
return l;
}
int mid = (l + r) >> 1;
pushdown(i, l, r, mid);
if (tr[i * 2].x > x) {
return Find(i * 2 + 1, mid + 1, r, x);
}
return Find(i * 2, l, mid, x);
}
void Solve() {
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> a[i];
}
build(1, 1, n);
mp.clear();
int ans = 0;
for (int i = 1; i <= n; i++) {
modify(1, 1, n, i, i, 0);
int k = Find(1, 1, n, a[i]);
if (k <= mp[a[i]]) {
modify(1, 1, n, k, mp[a[i]], a[i]);
}
mp[a[i]] = i;
ans += query(1, 1, n, 1, i);
}
cout << ans << "\n";
}
signed main() {
cin >> t;
while (t--) {
Solve();
}
return 0;
}
Not An SQRT Problem
说实话,就是更改一下 \(dfs\) 序即可,我们可以先更新儿子节点的 \(dfs\) 序,然后再递归
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 3e5 + 5, INF = 1e18;
struct node {
int x, lazy;
}tr[N * 4];
int n, q, dfn[N], cnt[N], sz[N], dcnt, first[N];
vector<int> g[N];
void dfs1(int u, int f) {
sz[u] = 1;
for (auto v : g[u]) {
if (v == f) {
continue;
}
dfs1(v, u);
sz[u] += sz[v];
}
}
void dfs2(int u, int f) {
for (auto v : g[u]) {
if (v == f) {
continue;
}
if (!first[u]) {
first[u] = v;
}
dfn[v] = ++dcnt;
cnt[u]++;
}
for (auto v : g[u]) {
if (v == f) {
continue;
}
dfs2(v, u);
}
}
void pushdown(int i) {
tr[i * 2].x += tr[i].lazy;
tr[i * 2].lazy += tr[i].lazy;
tr[i * 2 + 1].x += tr[i].lazy;
tr[i * 2 + 1].lazy += tr[i].lazy;
tr[i].lazy = 0;
}
node Merge(node x, node y) {
return {max(x.x, y.x), 0};
}
void modify(int i, int l, int r, int x, int y, int z) {
if (l > y || r < x) {
return ;
}
if (l >= x && r <= y) {
tr[i].x += z;
tr[i].lazy += z;
return ;
}
int mid = (l + r) >> 1;
pushdown(i);
modify(i * 2, l, mid, x, y, z);
modify(i * 2 + 1, mid + 1, r, x, y, z);
tr[i] = Merge(tr[i * 2], tr[i * 2 + 1]);
}
int query(int i, int l, int r, int x, int y) {
if (l > y || r < x) {
return -INF;
}
if (l >= x && r <= y) {
return tr[i].x;
}
int mid = (l + r) >> 1;
pushdown(i);
return max(query(i * 2, l, mid, x, y), query(i * 2 + 1, mid + 1, r, x, y));
}
signed main() {
ios::sync_with_stdio(0);
cin.tie(0);
cin >> n >> q;
for (int i = 1, u, v; i < n; i++) {
cin >> u >> v;
g[u].push_back(v);
g[v].push_back(u);
}
dfs1(1, 0);
dfn[1] = 1;
dcnt = 1;
dfs2(1, 0);
while (q--) {
int opt;
cin >> opt;
if (opt == 2) {
int x, v;
cin >> x >> v;
int cur = first[x];
modify(1, 1, n, dfn[cur], dfn[cur] + cnt[x] - 1, v);
}
else if (opt == 1) {
int x, v;
cin >> x >> v;
modify(1, 1, n, dfn[x], dfn[x], v);
if (!g[x].empty()) {
int cur = first[x];
modify(1, 1, n, dfn[cur], dfn[cur] + sz[x] - 2, v);
}
}
else if (opt == 3) {
int x;
cin >> x;
int ans = query(1, 1, n, dfn[x], dfn[x]);
if (!g[x].empty()) {
int cur = first[x];
ans = max(ans, query(1, 1, n, dfn[cur], dfn[cur] + sz[x] - 2));
}
cout << ans << '\n';
}
else {
int x;
cin >> x;
int cur = first[x];
cout << query(1, 1, n, dfn[cur], dfn[cur] + cnt[x] - 1) << "\n";
}
}
return 0;
}
01 on Tree
和一战到底或者魔塔一模一样
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 2e5 + 5;
struct node {
int u, a, b;
bool operator < (const node &_x) const {
return b * _x.a > _x.b * a;
}
};
int n, ans, fa[N], f[N], a[N], b[N];
priority_queue<node> q;
int find(int x) {
if (fa[x] == x) {
return x;
}
return fa[x] = find(fa[x]);
}
signed main() {
cin >> n;
for (int i = 2, u; i <= n; i++) {
cin >> u;
f[i] = u;
}
for (int i = 1, s; i <= n; i++) {
cin >> s;
a[i] = (s == 0);
b[i] = (s == 1);
q.push({i, (s == 0), (s == 1)});
fa[i] = i;
}
while (!q.empty()) {
if (q.top().u == 1 || a[q.top().u] != q.top().a || b[q.top().u] != q.top().b) {
q.pop();
continue;
}
int u = q.top().u, a1 = q.top().a, b1 = q.top().b;
q.pop();
int father = find(f[u]);
ans += b[father] * a1;
a[father] += a1;
b[father] += b1;
fa[u] = f[u];
q.push({father, a[father], b[father]});
}
cout << ans;
return 0;
}