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#include <algorithm>
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#include <iostream>
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#include <cstring>
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using namespace std;
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typedef long long LL;
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//快读
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LL read() {
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LL x = 0, f = 1;
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char ch = getchar();
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while (ch < '0' || ch > '9') {
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if (ch == '-') f = -1;
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ch = getchar();
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}
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while (ch >= '0' && ch <= '9') {
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x = (x << 3) + (x << 1) + (ch ^ 48);
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ch = getchar();
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}
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return x * f;
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}
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const int N = 200010;
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LL ans[N]; //每个询问的答案
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int n, m; //原始n个数字，共m个操作
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//线段树结构体
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#define ls u << 1
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#define rs u << 1 | 1
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struct Node {
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int l, r; //区间范围
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LL lazy; //区间修改懒标记
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LL sum; //区间和
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} tr[N << 2];
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//向上更新统计信息:区间和
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void pushup(int u) {
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tr[u].sum = tr[ls].sum + tr[rs].sum;
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}
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//向下传递区间修改标记
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void pushdown(int u) {
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//如果懒标记为0,没有可以传递
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if (!tr[u].lazy) return;
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//向下传递lazy标志
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tr[ls].lazy += tr[u].lazy, tr[rs].lazy += tr[u].lazy; //加法有叠加性
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//用lazy通过计算，更新左右儿子的sum值
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int l = tr[u].l, r = tr[u].r;
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int mid = (l + r) >> 1;
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tr[ls].sum += tr[u].lazy * (mid - l + 1);
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tr[rs].sum += tr[u].lazy * (r - mid);
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// lazy标记清零
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tr[u].lazy = 0;
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}
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//构建线段树
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void build(int u, int l, int r) {
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tr[u] = {l, r, 0, 0};
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if (l == r) return;
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int mid = (l + r) >> 1;
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build(ls, l, mid), build(rs, mid + 1, r);
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}
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//区间更新
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//之所以参数起名为[ql,qr]因为这是查询区间，而不是构建区间
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void modify(int u, int ql, int qr, LL c) {
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int l = tr[u].l, r = tr[u].r; //将变量名l,r保留下来给tr[u].l,tr[u].r
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if (ql <= l && qr >= r) { // u结点的管辖范围[tr[u].l~tr[u].r]完整命中
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tr[u].lazy += c; //区间修改加上lazy标记,这样就不用现在就逐层下传了
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tr[u].sum += c * (r - l + 1); //对统计信息进行更新,完成本层的统计信息更新
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return;
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}
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//下面要进行分裂，分裂前需要pushdown
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pushdown(u);
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int mid = (l + r) >> 1;
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if (ql <= mid) modify(ls, ql, qr, c); //与左儿子区间有交集，递归左儿子
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if (mid < qr) modify(rs, ql, qr, c); //与右儿子区间有交集，递归右儿子
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//向上更新统计信息
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pushup(u);
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}
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//查询区间sum和
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LL query(int u, int ql, int qr) {
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int l = tr[u].l, r = tr[u].r;
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if (ql <= l && r <= qr) return tr[u].sum; //完整命中，返回区间和
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LL res = 0;
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//分裂前需要pushdown
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pushdown(u);
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int mid = (l + r) >> 1;
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if (ql <= mid) res += query(ls, ql, qr); //与左儿子区间有交集，递归左儿子
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if (mid < qr) res += query(rs, ql, qr); //与右儿子区间有交集，递归右儿子
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return res;
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}
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//整体二分的查询结构体
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struct Q {
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int op, l, r;
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LL k;
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int id;
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} q[N], q1[N], q2[N];
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//整体二分
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void solve(int l, int r, int ql, int qr) {
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if (ql > qr) return; //无效查询区间，防RE
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if (l == r) {
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for (int i = ql; i <= qr; i++)
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if (q[i].op == 2) ans[q[i].id] = l; //二分标识答案
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return;
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}
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int mid = (l + r) >> 1;
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int nl = 0, nr = 0;
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for (int i = ql; i <= qr; i++) {
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if (q[i].op == 1) { //增加
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if (q[i].k > mid) {
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modify(1, q[i].l, q[i].r, 1); //用线段树指定区全都+1
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q1[++nl] = q[i]; //加入左增加操作
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} else
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q2[++nr] = q[i]; //加入右增加操作
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} else { //查询
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LL t = query(1, q[i].l, q[i].r);
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if (t >= q[i].k)
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q1[++nl] = q[i];
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else {
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q[i].k -= t;
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q2[++nr] = q[i];
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}
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}
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}
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//有加就有减
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for (int i = ql; i <= qr; i++)
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if (q[i].op == 1)
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if (q[i].k > mid) modify(1, q[i].l, q[i].r, -1);
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//合并到q数组
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for (int i = ql; i < ql + nl; i++) q[i] = q1[i - ql + 1];
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for (int i = ql + nl; i <= qr; i++) q[i] = q2[i - ql - nl + 1];
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//递归左右
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solve(mid + 1, r, ql, ql + nl - 1), solve(l, mid, ql + nl, qr);
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}
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int main() {
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//原序列n个数字，共m个操作
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n = read(), m = read();
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//构建线段树
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build(1, 1, n);
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int qc = 0; //查询的个数
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for (int i = 1; i <= m; i++) {
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int op = read(), l = read(), r = read(), k = read();
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if (op == 2) //查询操作
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q[i] = {op, l, r, k, ++qc};
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else // 增加操作 op=1
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// 1 l r k : op=1 增加，[l,r]:增加k
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q[i] = {op, l, r, k};
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}
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//整体二分
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solve(-1e9, 1e9, 1, m);
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//输出
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for (int i = 1; i <= qc; i++) printf("%lld\n", ans[i]);
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return 0;
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} |
