数据结构 【树状数组】【线段树】【珂朵莉树】
2024-04-25 05:28:28
一.区间合并
1.AcWing245你能回答这些问题吗
#include <bits/stdc++.h>using namespace std;const int inf = 0x3f3f3f3f;
int n, m;
int ar[500050];
struct node
{int l, r;int mx, mx_l, mx_r, sum;
}tree[2000050];
int op, x, y;void pushup(node &a, node &b, node &c)
{a.mx = max(max(b.mx, c.mx), b.mx_r + c.mx_l);a.mx_l = max(b.mx_l, b.sum + c.mx_l);a.mx_r = max(c.mx_r, c.sum + b.mx_r);a.sum = b.sum + c.sum;
}void build(int p, int l, int r)
{tree[p].l = l, tree[p].r = r;if(l == r){tree[p].mx = tree[p].mx_l = tree[p].mx_r = tree[p].sum = ar[l];return ;}int mid = (l + r) >> 1;build(p<<1, l, mid);build(p<<1|1, mid + 1, r);pushup(tree[p], tree[p<<1], tree[p<<1|1]);
}node query(int p, int l, int r)
{if(l <= tree[p].l && tree[p].r <= r) return tree[p];int mid = (tree[p].l + tree[p].r) >> 1;if(r <= mid) return query(p<<1, l, r);else if(l > mid) return query(p<<1|1, l, r);else{node left = query(p<<1, l, r);node right = query(p<<1|1, l, r);node res;pushup(res, left, right);return res;}
}void updata(int p, int x, int y)
{if(tree[p].l == tree[p].r) tree[p].mx = tree[p].mx_l = tree[p].mx_r = tree[p].sum = y;else{int mid = (tree[p].l + tree[p].r) >> 1;if(mid >= x) updata(p<<1, x, y);else updata(p<<1|1, x, y);pushup(tree[p], tree[p<<1], tree[p<<1|1]);}
}int main()
{scanf("%d%d", &n, &m);for(int i = 1; i <= n; ++i) scanf("%d", &ar[i]);build(1, 1, n);while(m--){scanf("%d%d%d", &op, &x, &y);if(op == 1){if(x > y) swap(x, y);printf("%d\n", query(1, x, y).mx);}else updata(1, x, y);}return 0;
}2.Codeforces Round #742 (Div. 2) E. Non-Decreasing Dilemma
同上一个题一样，除了维护询问的属性sum以外,再维护子数组左右边界分别为区间边界时最长的长度len_l,len_r,，以及区间长度len,即可实现区间合并。
#include <bits/stdc++.h>using namespace std;
typedef long long ll;#define io ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
int n, q;
ll ar[200050];
struct node
{int l, r;ll len_l, len_r, sum, len;// void print()
// {// cout << l << ' ' << r << ' ' << len_l << ' ' << len_r << ' ' << sum << ' ' << len << '\n';
// }
}tree[800050];
int op;
ll x, y;ll calc(ll n)
{return (1ll + n) * n / 2ll;
}void pushup(node &p, node &le, node &ri)
{p.l = le.l;p.r = ri.r;if(ar[le.r] <= ar[ri.l]){p.sum = le.sum + ri.sum - calc(le.len_r) - calc(ri.len_l) + calc(le.len_r + ri.len_l);if(le.len == le.len_l) p.len_l = le.len + ri.len_l;else p.len_l = le.len_l;if(ri.len_r == ri.len) p.len_r = ri.len + le.len_r;else p.len_r = ri.len_r;p.len = le.len + ri.len;}else{p.sum = le.sum + ri.sum;p.len_l = le.len_l;p.len_r = ri.len_r;p.len = le.len + ri.len;}
}void pushup(int p)
{pushup(tree[p], tree[p<<1], tree[p<<1|1]);
}void build(int p, int l, int r)
{tree[p].l = l, tree[p].r = r;if(l == r){tree[p].len_l = tree[p].len_r = tree[p].sum = tree[p].len = 1;return ;}int mid = (l + r) >> 1;build(p<<1, l, mid);build(p<<1|1, mid + 1, r);pushup(p);
}void updata(int p, int pos, ll y)
{if(tree[p].l == tree[p].r) ar[pos] = y;else{int mid = (tree[p].l + tree[p].r) >> 1;if(mid >= x) updata(p<<1, pos, y);else updata(p<<1|1, pos, y);pushup(p);}
}node query(int p, int l, int r)
{if(l <= tree[p].l && tree[p].r <= r){//tree[p].print();return tree[p];}int mid = (tree[p].l + tree[p].r) >> 1;if(r <= mid) return query(p<<1, l, r);else if(l > mid) return query(p<<1|1, l, r);else{node res;node left = query(p<<1, l, r);node right = query(p<<1|1, l, r);pushup(res, left, right);//res.print();return res;}}int main()
{cin >> n >> q;for(int i = 1; i <= n; ++i) cin >> ar[i];build(1, 1, n);// for(int i = 1; i <= n * 4; ++i)
// {// cout << i << ' ' ;
// tree[i].print();
// }while(q--){cin >> op >> x >> y;if(op == 1) updata(1, x, y);else{//query(1, x, y).print();cout << query(1, x, y).sum << '\n';}}return 0;
}二.珂朵莉树
牛客练习赛100 E 小红的公倍数(珂朵莉树)
#include <bits/stdc++.h>using namespace std;
typedef long long ll;
#define int ll
typedef vector<int> vv;
#define itt set<node>::iterator
#define io ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)const int maxn = 20005;
const ll mod = 1e9 + 7;
int n, q, l, r, num;
ll ar[maxn];
struct node
{int l, r;vv vt;bool operator < (const node &b) const{return l < b.l;}
};
set<node> tree;
bool prime[maxn];
int Prime[maxn];
int pid[maxn];void make_prime()
{memset(prime, true, sizeof(prime));memset(pid, -1, sizeof(pid));prime[0] = prime[1] = false;for(int i = 2; i <= maxn; i++){if(prime[i]){Prime[num]=i;pid[i] = num++;}for(int j=0; j < num && i * Prime[j] < maxn; j++){prime[i * Prime[j]] = false;if(!(i % Prime[j]))break;}}
}ll pow_mod(ll a, ll n)
{ll ans = 1;a %= mod;while(n){if(n & 1) ans = (ans * a) % mod;a = (a * a) % mod;n >>= 1;}return ans;
}itt split(int pos)
{itt it = tree.lower_bound({pos, 0, {}});if(it != tree.end() && it -> l == pos) return it;--it;int l = it -> l;int r = it -> r;vector<int> v = it -> vt;tree.erase(it);tree.insert({l, pos - 1, v});return tree.insert({pos, r, v}).first;
}void solve(int l, int r)
{itt itr = split(r + 1);itt itl = split(l);
// cout << itl -> l << ' ' << itl -> r << '\n';
// cout << itr -> l << ' ' << itr -> r << '\n';vector<int> vt(2300, 0);for(itt it = itl; it != itr; ++it){//cout << it -> l << ' ' << it -> t << '\n';for(int i = 0; i < 2300; ++i){vt[i] = max(vt[i], it->vt[i]);//cout << vt[i] <<}}ll ans = 1;for(int i = 0; i < 2300; ++i){//cout << i << ' ' << vt[i] << ' ';ans = (ans * pow_mod(Prime[i], vt[i])) % mod;//cout << ans << '\n';}cout << ans << '\n';tree.erase(itl, itr);tree.insert(node{l, r, vt});
}void print()
{for(itt it = tree.begin(); it != tree.end(); ++it){cout << it -> l << ' ' << it -> r << '\n';}
}signed main()
{io;make_prime();
// for(int i = 0; i <= 100; ++i)
// {// cout << i << ' ' << prime[i] << ' ' << Prime[i] << ' ' << pid[i] << '\n';
// }cin >> n >> q;vector<int> vt(2300, 0);int cnt = 0;for(int i = 1; i <= n; ++i){cin >> ar[i];vt.clear();vt = vector<int>(2300, 0);for(int j = 2; j * j <= ar[i]; ++j){cnt = 0;while(ar[i] % j == 0){++cnt;ar[i] /= j;}if(pid[j] == -1) continue;vt[pid[j]] += cnt;}if(ar[i] > 1) ++vt[pid[ar[i]]];tree.insert(node{i, i, vt});}while(q--){cin >> l >> r;solve(l, r);//print();}return 0;
}三.区间操作
1.2022 ACM 湖北省赛 L (线段树)
#include <bits/stdc++.h>using namespace std;
typedef long long ll;
#define int ll
#define io ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)const int maxn = 1e5 + 5;
int n, m;
int ar[maxn];
bool s[maxn];
struct node
{int l, r;int lazy_sum;int cnt, sum;int mid(){return (l + r) >> 1;}void print(){cout << l << ' ' << r << ' ' << lazy_sum << ' ' << cnt << ' ' << sum << '\n';}
} tree[maxn<<2];
int op, l, r, x;void pushup(int p)
{tree[p].sum = tree[p<<1].sum + tree[p<<1|1].sum;tree[p].cnt = tree[p<<1].cnt + tree[p<<1|1].cnt;
}void pushdown(int p)
{tree[p<<1].sum += tree[p<<1].cnt * tree[p].lazy_sum;tree[p<<1].lazy_sum += tree[p].lazy_sum;tree[p<<1|1].sum += tree[p<<1|1].cnt * tree[p].lazy_sum;tree[p<<1|1].lazy_sum += tree[p].lazy_sum;tree[p].lazy_sum = 0;
}void build(int p, int l, int r)
{tree[p].l = l;tree[p].r = r;tree[p].lazy_sum = 0;if(l == r){tree[p].sum = ar[l];tree[p].cnt = s[l];return ;}int mid = (l + r) >> 1;build(p<<1, l, mid);build(p<<1|1, mid + 1, r);pushup(p);
}void updata1(int p, int pos)
{if(tree[p].l == tree[p].r){tree[p].cnt ^= 1;return ;}if(tree[p].lazy_sum) pushdown(p);int mid = (tree[p].l + tree[p].r) >> 1;if(pos <= mid) updata1(p<<1, pos);else updata1(p<<1|1, pos);pushup(p);
}void updata2(int p, int l, int r, int x)
{if(l <= tree[p].l && tree[p].r <= r){tree[p].sum += tree[p].cnt * x;tree[p].lazy_sum += x;return ;}if(tree[p].lazy_sum) pushdown(p);int mid = (tree[p].l + tree[p].r) >> 1;if(l <= mid) updata2(p<<1, l, r, x);if(r > mid) updata2(p<<1|1, l, r, x);pushup(p);
}ll query(int p, int l, int r)
{if(l <= tree[p].l && tree[p].r <= r) return tree[p].sum;if(tree[p].lazy_sum) pushdown(p);int mid = (tree[p].l + tree[p].r) >> 1;ll res = 0;if(l <= mid) res = query(p<<1, l, r);if(r > mid) res += query(p<<1|1, l, r);return res;
}signed main()
{io;cin >> n >> m;for(int i = 1; i <= n; ++i) cin >> ar[i];for(int i = 1; i <= n; ++i) cin >> s[i];build(1, 1, n);// for(int i = 1; i <= 3 * n; ++i)
// {// cout << i << ' ';
// tree[i].print();
// }while(m--){cin >> op;if(op == 1 || op == 2){cin >> x;updata1(1, x);
// for(int i = 1; i <= 7; ++i)
// {// cout << i << ' ';
// tree[i].print();
// }}else if(op == 3){cin >> l >> r >> x;updata2(1, l, r, x);
// for(int i = 1; i <= 7; ++i)
// {// cout << i << ' ';
// tree[i].print();
// }}else{cin >> l >> r;cout << query(1, l, r) << '\n';
// for(int i = 1; i <= 7; ++i)
// {// cout << i << ' ';
// tree[i].print();
// }}}return 0;
}2.2022牛客寒假算法基础集训营4 B-进制 (线段树)
分析：
维护区间最大值，和作为i进制数的大小即可维护。
AC代码：
#include <bits/stdc++.h>using namespace std;
typedef long long ll;
#define int ll
#define io ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define endl '\n'const int maxn = 1e5 + 5;
const ll mod = 1e9 + 7;
const ll inf = 0x3f3f3f3f;
struct node
{int l, r, mx;int ar[12];int mid(){return (l + r) >> 1;}void print(){cout << l << ' ' << r << ' ' << mx << '\n';for(int i = 1; i <= 10; ++i) cout << ar[i] << ' ';cout << '\n';}
}tree[maxn<<2];
int n, q, op, x, y;
string s;
node ans;ll pow_mod(ll a, ll n)
{ll ans = 1;a %= mod;while(n){if(n & 1) ans = (ans * a) % mod;a = (a * a) % mod;n >>= 1;}return ans;
}void pushup(node &p, node &le, node &ri)
{p.mx = max(le.mx, ri.mx);p.l = le.l;p.r = ri.r;for(int i = 1; i <= 10; ++i){if(i <= p.mx) p.ar[i] = inf;else p.ar[i] = ((le.ar[i] * pow_mod(i, ri.r - ri.l + 1)) % mod + ri.ar[i]) % mod;}
}void pushup(int p)
{pushup(tree[p], tree[p<<1], tree[p<<1|1]);
}void build(int p, int l, int r)
{tree[p].l = l;tree[p].r = r;if(l == r){tree[p].mx = s[l] - '0';for(int i = 1; i <= 10; ++i){if(i <= tree[p].mx) tree[p].ar[i] = inf;else tree[p].ar[i] = tree[p].mx;}return ;}int mid = (l + r) >> 1;build(p<<1, l, mid);build(p<<1|1, mid + 1, r);pushup(p);
}void updata(int p, int x, int y)
{if(tree[p].l == tree[p].r){tree[p].mx = y;for(int i = 1; i <= 10; ++i){if(i <= tree[p].mx) tree[p].ar[i] = inf;else tree[p].ar[i] = tree[p].mx;}return ;}int mid = tree[p].mid();if(x <= mid) updata(p<<1, x, y);else updata(p<<1|1, x, y);pushup(p);
}node query(int p, int l, int r)
{if(l <= tree[p].l && tree[p].r <= r) return tree[p];int mid = tree[p].mid();if(mid >= r) return query(p<<1, l, r);else if(mid < l) return query(p<<1|1, l, r);else{node res;node le = query(p<<1, l, r);node ri = query(p<<1|1, l, r);pushup(res, le, ri);return res;}
}signed main()
{io;cin >> n >> q;cin >> s;s = " " + s;build(1, 1, n);// for(int i = 1; i <= 9; ++i)
// {// cout << i << ' ';
// tree[i].print();
// }while(q--){cin >> op >> x >> y;if(op == 1) updata(1, x, y);else{ans = query(1, x, y);cout << ans.ar[ans.mx + 1] << '\n';}}return 0;
}3.Codeforces Round #254 (Div. 1) C. DZY Loves Colors (线段树)
分析：
维护区间权值和，区间的颜色，如果区间内颜色统一则区间修改，使用lazy标记，不统一则暴力单点修改。
AC代码：
#include <bits/stdc++.h>using namespace std;
typedef long long ll;
#define int ll
#define io ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define endl '\n'
#define itt set<node>::iteratorconst int maxn = 1e5 + 5;
const ll mod = 1e9 + 7;
const ll inf = 0x3f3f3f3f;
struct node
{int l, r;int color, sum, cnt;int lazy;bool flag;int mid(){return (l + r) >> 1;}void print(){cout << l << ' ' << r << ' ' << color << ' ' << flag << ' ' << sum << ' '<< cnt << ' ' << lazy << '\n';}
}tree[maxn<<2];
int n, m, op, l, r, x;void pushup(int p)
{tree[p].sum = tree[p<<1].sum + tree[p<<1|1].sum;tree[p].flag = false;tree[p].color = 0;if(tree[p<<1].flag && tree[p<<1|1].flag && tree[p<<1].color == tree[p<<1|1].color){tree[p].flag = true;tree[p].color = tree[p<<1].color;}
}void pushdown(int p)
{tree[p<<1].sum += tree[p<<1].cnt * tree[p].lazy;tree[p<<1].lazy += tree[p].lazy;tree[p<<1].color = tree[p].color;tree[p<<1|1].sum += tree[p<<1|1].cnt * tree[p].lazy;tree[p<<1|1].lazy += tree[p].lazy;tree[p<<1|1].color = tree[p].color;tree[p].lazy = 0;
}void build(int p, int l, int r)
{tree[p].l = l;tree[p].r = r;tree[p].cnt = r - l + 1;tree[p].lazy = 0;if(l == r){tree[p].color = l;tree[p].flag = true;tree[p].sum = 0;return ;}int mid = tree[p].mid();build(p<<1, l, mid);build(p<<1|1, mid + 1, r);pushup(p);
}void updata(int p, int l, int r, int x)
{if (tree[p].l > r || tree[p].r < l) return;if(l <= tree[p].l && tree[p].r <= r && tree[p].flag){tree[p].sum += tree[p].cnt * abs(tree[p].color - x);tree[p].lazy += abs(tree[p].color - x);tree[p].color = x;return ;}if(tree[p].lazy) pushdown(p);updata(p<<1, l, r, x);updata(p<<1|1, l, r, x);pushup(p);
}ll query(int p, int l, int r)
{if(tree[p].l > r || tree[p].r < l) return 0;if(l <= tree[p].l && tree[p].r <= r) return tree[p].sum;if(tree[p].lazy) pushdown(p);return query(p<<1, l, r) + query(p<<1|1, l, r);
}signed main()
{io;cin >> n >> m;build(1, 1, n);while(m--){cin >> op;if(op == 1){cin >> l >> r >> x;updata(1, l, r, x);}else{cin >> l >> r;cout << query(1, l, r) << '\n';}}return 0;
}四.推结论+数据结构
1.同济大学程序设计竞赛 G(离线+树状数组)
#include <bits/stdc++.h>using namespace std;
typedef long long ll;
#define int ll
#define io ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define endl '\n'
#define itt set<node>::iterator
#define pii pair<int,int>const int maxn = 4e5 + 5;
const ll mod = 1e9 + 7;
const ll inf = 0x3f3f3f3f;
int ar[maxn], pre[maxn];
struct node
{int k, x, id;
};
vector<node> vt[maxn];
int ans[maxn], ans_1[maxn], ans_2[maxn], ans_3[maxn];
//答案，区间内小于等于k/2的数字之和，区间内大于k/2的数字个数，区间内大于k/2的数字之和
vector<int> vv;
int tree[maxn];
int sum[maxn];
int t, n, q, l, r, h;
struct node1
{int l, r, k, id;
}ask[maxn];int lowbit(int x)
{return (-x) & x;
}void updata(int pos, int x)
{while(pos <= n + q){++tree[pos];sum[pos] += x;pos += lowbit(pos);}
}pii get_sum(int pos)
{pii pi;pi.first = pi.second = 0;while(pos){pi.first += tree[pos];pi.second += sum[pos];pos -= lowbit(pos);}return pi;
}void work()
{cin >> n >> q;for(int i = 1; i <= n; ++i){cin >> ar[i];pre[i] = pre[i - 1] + ar[i];vv.push_back(ar[i]);}for(int i = 1; i <= q; ++i){cin >> l >> r >> h;ask[i].l = l;ask[i].r = r;ask[i].k = h;ask[i].id = i;vv.push_back(h / 2);vt[l - 1].push_back({-1, h / 2, i});vt[r].push_back({1, h / 2, i});}sort(vv.begin(), vv.end());vv.erase(unique(vv.begin(), vv.end()), vv.end());int id, pos;pii pi;for(int i = 1; i <= n; ++i){pos = lower_bound(vv.begin(), vv.end(), ar[i]) - vv.begin() + 1;updata(pos, ar[i]);for(int j = 0; j < vt[i].size(); ++j){id = vt[i][j].id;pos = lower_bound(vv.begin(), vv.end(), vt[i][j].x) - vv.begin() + 1;pi = get_sum(pos);ans_1[id] += vt[i][j].k * pi.second;ans_2[id] += vt[i][j].k * pi.first;}}for(int i = 1; i <= q; ++i){ans[i] = ans_1[i];ans_2[i] = (ask[i].r - ask[i].l + 1) - ans_2[i];ans_2[i] = ans_2[i] * ask[i].k;ans_3[i] = pre[ask[i].r] - pre[ask[i].l - 1] - ans_1[i];ans[i] += ans_2[i] - ans_3[i];cout << ans[i] << '\n';}
}signed main()
{//io;work();return 0;
}2.daimayuan 喵 ~ 喵~ 喵 ~
#include <bits/stdc++.h>using namespace std;
typedef long long ll;
#define int ll
#define io ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define endl '\n'
#define itt set<node>::iterator
#define pii pair<int,int>const int maxn = 4e5 + 5;
const ll mod = 1e9 + 7;
const ll inf = 0x3f3f3f3f;
int n, m, op, l, r;
int sum1[maxn], sum2[maxn];int lowbit(int x)
{return (-x) & x;
}void updata1(int pos)
{while(pos <= n){++sum1[pos];pos += lowbit(pos);}
}void updata2(int pos)
{while(pos <= n){++sum2[pos];pos += lowbit(pos);}
}int get_sum1(int pos)
{int res = 0;while(pos){res += sum1[pos];pos -= lowbit(pos);}return res;
}int get_sum2(int pos)
{int res = 0;while(pos){res += sum2[pos];pos -= lowbit(pos);}return res;
}signed main()
{io;cin >> n >> m;while(m--){cin >> op >> l >> r;if(op == 1){updata1(l);updata2(r);}else{cout << get_sum1(r) - get_sum2(l - 1) << '\n';}}return 0;
}3.Educational Codeforces Round 126 (Rated for Div. 2) D. Progressions Covering
#include <bits/stdc++.h>using namespace std;
typedef long long ll;
#define int ll
#define io ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define endl '\n'
#define itt set<node>::iterator
#define pii pair<int,int>const int maxn = 3e5 + 5;
const ll mod = 1e9 + 7;
const ll inf = 0x3f3f3f3f;
int n, k;
int ar[maxn];
struct node
{int l, r;int sum, lazy, len;int mid(){return (l + r) >> 1;}void print(){cout << l << ' ' << r << ' ' << sum << '\n';}
}tree[maxn<<2];void build(int p, int l, int r)
{tree[p].l = l;tree[p].r = r;tree[p].len = r - l + 1;tree[p].lazy = tree[p].sum = 0;if(l == r) return ;int mid = tree[p].mid();build(p<<1, l, mid);build(p<<1|1, mid + 1, r);
}void pushdown(int p)
{tree[p<<1].sum += tree[p].lazy * tree[p<<1].len;tree[p<<1].lazy += tree[p].lazy;tree[p<<1|1].sum += tree[p].lazy * tree[p<<1|1].len;tree[p<<1|1].lazy += tree[p].lazy;tree[p].lazy = 0;
}void pushup(int p)
{tree[p].sum = tree[p<<1].sum + tree[p<<1|1].sum;
}void updata(int p, int l, int r, int x)
{if(l <= tree[p].l && tree[p].r <= r){tree[p].sum += tree[p].len * x;tree[p].lazy += x;return ;}int mid = tree[p].mid();if(tree[p].lazy) pushdown(p);if(l <= mid) updata(p<<1, l, r, x);if(r > mid) updata(p<<1|1, l, r, x);pushup(p);
}int query(int p, int l, int r)
{if(l <= tree[p].l && tree[p].r <= r) return tree[p].sum;int mid = tree[p].mid();int res = 0;if(tree[p].lazy) pushdown(p);if(l <= mid) res += query(p<<1, l, r);if(r > mid) res += query(p<<1|1, l, r);return res;
}signed main()
{io;cin >> n >> k;for(int i = 1; i <= n; ++i) cin >> ar[i];build(1, 1, n);int sub, mx, cnt, ans = 0;for(int i = n; i > 0; --i){sub = query(1, 1, i);ar[i] -= sub;if(ar[i] <= 0) continue;mx = min(i, k);cnt = (ar[i] + mx - 1) / mx;ans += cnt;updata(1, i - mx + 1, i, cnt);}cout << ans << '\n';return 0;
}
#include <bits/stdc++.h>using namespace std;
typedef long long ll;
#define int ll
#define io ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define endl '\n'
#define itt set<node>::iterator
#define pii pair<int,int>const int maxn = 4e5 + 5;
const ll mod = 1e9 + 7;
const ll inf = 0x3f3f3f3f;
int n, k, cnt, len;
int ar[maxn];
queue<pii> q;
int ans;signed main()
{io;cin >> n >> k;for(int i = 1; i <= n; ++i) cin >> ar[i];int sum = 0, sum_cnt = 0;for(int i = n; i > 0; --i){sum -= sum_cnt;ar[i] -= sum;if(ar[i] > 0){len = min(i, k);cnt = (ar[i] + len - 1ll) / len;q.push({i - len + 1ll, cnt});sum += len * cnt;sum_cnt += cnt;ans += cnt;}if(q.front().first == i){sum -= q.front().second;sum_cnt -= q.front().second;q.pop();}}cout << ans << '\n';return 0;
}五.dp/思维+数据结构
1.最长上升子序列计数（Bonus）(DP/线段树/树状数组)
#include <bits/stdc++.h>using namespace std;
typedef long long ll;
#define int ll
#define io ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define endl '\n'
#define itt set<node>::iterator
#define pii pair<int,int>const int maxn = 4e5 + 5;
const ll mod = 1e9 + 7;
const ll inf = 0x3f3f3f3f;
int n;
int ar[maxn], dp[maxn], sum[maxn];
vector<int> vt;
int ans;
struct node
{int l, r;int mx_len, sum;int mid(){return (l + r) >> 1;}void print(){cout << l << ' ' << r << ' ' << mx_len << ' ' << sum << '\n';}
} tree[maxn<<2];void pushup(node &p, node &le, node &ri)
{p.l = le.l;p.r = ri.r;p.mx_len = max(le.mx_len, ri.mx_len);p.sum = 0;if(le.mx_len == p.mx_len) p.sum = (p.sum + le.sum) % mod;if(ri.mx_len == p.mx_len) p.sum = (p.sum + ri.sum) % mod;
}void pushup(int p)
{pushup(tree[p], tree[p<<1], tree[p<<1|1]);
}void build(int p, int l, int r)
{tree[p].l = l;tree[p].r = r;tree[p].mx_len = tree[p].sum = 0;if(l == r) return ;int mid = tree[p].mid();build(p<<1, l, mid);build(p<<1|1, mid + 1, r);
}void updata(int p, int pos, int len, int num)
{if(tree[p].l == tree[p].r){tree[p].mx_len = len;tree[p].sum = num;return ;}int mid = tree[p].mid();if(pos <= mid) updata(p<<1, pos, len, num);else updata(p<<1|1, pos, len, num);pushup(p);
}node query(int p, int l, int r)
{if(l > r) return {0, 0, 0, 0};if(l <= tree[p].l && tree[p].r <= r) return tree[p];int mid = tree[p].mid();if(r <= mid) return query(p<<1, l, r);else if(l > mid) return query(p<<1|1, l, r);else{node res;node le = query(p<<1, l, r);node ri = query(p<<1|1, l, r);pushup(res, le, ri);return res;}
}signed main()
{io;cin >> n;for(int i = 1; i <= n; ++i){cin >> ar[i];vt.push_back(ar[i]);}build(1, 1, n);sort(vt.begin(), vt.end());vt.erase(unique(vt.begin(), vt.end()), vt.end());int pos;node pp;for(int i = 1; i <= n; ++i){pos = lower_bound(vt.begin(), vt.end(), ar[i]) - vt.begin() + 1;dp[pos] = max(1ll, dp[pos]);sum[pos] = max(0ll, sum[pos]);pp = query(1, 1, pos - 1);if(pp.mx_len + 1 == 1){dp[pos] = 1;++sum[pos];}else if(dp[pos] < pp.mx_len + 1){dp[pos] = pp.mx_len + 1;sum[pos] = pp.sum;}else if(dp[pos] == pp.mx_len + 1) sum[pos] = (sum[pos] + pp.sum) % mod;updata(1, pos, dp[pos], sum[pos]);}cout << query(1, 1, n).sum << '\n';return 0;
}2.Codeforces Round #783 (Div. 2) D. Optimal Partition 【dp+线段树】
#include <bits/stdc++.h>using namespace std;
typedef long long ll;
#define int ll
#define io ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define endl '\n'
#define itt set<node>::iterator
#define pii pair<int,int>const int inf = 0x3f3f3f3f;
const int maxn = 5e5 + 5;
int t, n, x;
int sum[maxn];
int dp[maxn];
vector<int> vt;
struct node
{int l, r;int dp, dp1, dp2;int mid(){return (l + r) >> 1;}void print(){cout << l << ' ' << r << ' ' << dp << ' ' << dp1 << ' ' << dp2 << '\n';}
} tree[maxn<<2];void pushup(node &p, node &le, node &ri)
{p.l = le.l;p.r = ri.r;p.dp = max(le.dp, ri.dp);p.dp1 = max(le.dp1, ri.dp1);p.dp2 = max(le.dp2, ri.dp2);
}void pushup(int p)
{pushup(tree[p], tree[p<<1], tree[p<<1|1]);
}void build(int p, int l, int r)
{tree[p].l = l;tree[p].r = r;if(l == r){tree[p].dp = tree[p].dp1 = tree[p].dp2 = -inf;return ;}int mid = tree[p].mid();build(p<<1, l, mid);build(p<<1|1, mid + 1, r);pushup(p);
}void updata(int p, int pos, int dp, int id)
{if(tree[p].l == tree[p].r){tree[p].dp = max(tree[p].dp, dp);tree[p].dp1 = max(tree[p].dp1, dp - id);tree[p].dp2 = max(tree[p].dp2, dp + id);return ;}int mid = tree[p].mid();if(pos <= mid) updata(p<<1, pos, dp, id);else updata(p<<1|1, pos, dp, id);pushup(p);
}node query(int p, int l, int r)
{if(l > r) return {-inf, -inf, -inf, -inf};if(l <= tree[p].l && tree[p].r <= r) return tree[p];int mid = tree[p].mid();if(mid >= r) return query(p<<1, l, r);else if(mid < l) return query(p<<1|1, l, r);else{node le = query(p<<1, l, r);node ri = query(p<<1|1, l, r);node res;pushup(res, le, ri);return res;}
}signed main()
{io;cin >> t;while(t--){vt.clear();cin >> n;for(int i = 1; i <= n; ++i){cin >> x;sum[i] = sum[i - 1] + x;vt.push_back(sum[i]);dp[i] = 0;}vt.push_back(0);sort(vt.begin(), vt.end());vt.erase(unique(vt.begin(), vt.end()), vt.end());build(1, 1, vt.size());node pp;int mx1, mx2, mx3, pos;pos = lower_bound(vt.begin(), vt.end(), 0) - vt.begin() + 1;updata(1, pos, 0, 0);for(int i = 1; i <= n; ++i){pos = lower_bound(vt.begin(), vt.end(), sum[i]) - vt.begin() + 1;mx1 = query(1, 1, pos - 1).dp1;mx2 = query(1, pos, pos).dp;mx3 = query(1, pos + 1, vt.size()).dp2;dp[i] = max(mx1 + i, max(mx2, mx3 - i));updata(1, pos, dp[i], i);}cout << dp[n] << '\n';}return 0;
}3.Codeforces Round #609 (Div. 1) C. K Integers
#include <bits/stdc++.h>using namespace std;
typedef long long ll;
#define int ll
#define io ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define endl '\n'
#define itt set<node>::iterator
#define pii pair<int,int>const int inf = 0x3f3f3f3f;
const int maxn = 2e5 + 5;
int t, n, x;
int ar[maxn];
int pos[maxn];
int ans[maxn];
int tree1[maxn], tree2[maxn];int lowbit(int x)
{return (-x) & x;
}void updata(int x)
{int i = x;while(i <= n){tree1[i]++;tree2[i] += x;i += lowbit(i);}
}int get_cnt(int x)
{int res = 0;while(x){res += tree1[x];x -= lowbit(x);}return res;
}int get_sum(int x)
{int res = 0;while(x){res += tree2[x];x -= lowbit(x);}return res;
}int binary_find(int x)
{int r = n, l = 1;while(l <= r){int mid = (l + r) >> 1;if(get_cnt(mid) >= x) r = mid - 1;else l = mid + 1;}return r + 1;
}signed main()
{io;cin >> n;for(int i = 1; i <= n; ++i){cin >> ar[i];pos[ar[i]] = i;}ans[1] = 0;updata(pos[1]);int mid, cnt, sum, pre = 0;for(int i = 2; i <= n; ++i){ans[i] += pre;ans[i] += get_cnt(n) - get_cnt(pos[i]);pre = ans[i];updata(pos[i]);cnt = (i + 1ll) / 2ll;mid = binary_find(cnt);sum = get_sum(mid);ans[i] += cnt * mid - cnt * (cnt - 1ll) / 2ll - sum;cnt = i - cnt;sum = get_sum(n) - get_sum(mid);ans[i] += sum - cnt * (mid + 1ll) - cnt * (cnt - 1ll) / 2ll;}for(int i = 1; i <= n; ++i) cout << ans[i] << ' ';cout << '\n';return 0;
}
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