P1198 [JSOI2008]最大数

st 表 和 分块 都可做 还跑得很快

然而我只会线段树

题目要求是线段树的插入操作

直接考虑建一棵足够大的线段树，插入操作就变成了 单点修改操作

剩下的查询直接查询$ (x \ - \ L \ + \ 1, \ x )$

code:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 200001;
const ll INF = 99999999999999;
int n;
ll D, t = 0, a[N], cnt = 0;

struct Stree {
    int l, r; ll maxi; 
}tree[N * 3];    

void pushup(int rt) {
    tree[rt].maxi = max(tree[rt << 1].maxi, tree[rt << 1 | 1].maxi);
}

void build(int l, int r, int rt) {
    if(l == r) {
        tree[rt].maxi = -INF; return;
    }
    int mid = (l + r) >> 1;
    build(l, mid, rt << 1), build(mid + 1, r, rt << 1 | 1);
    pushup(rt);
}

void upd(int idx, int C, int l, int r, int rt) {
    if(l == r) { tree[rt].maxi = C; return; }
    int mid = (l + r) >> 1;
    if(idx <= mid) upd(idx, C, l, mid, rt << 1);
    else upd(idx, C, mid + 1, r, rt << 1 | 1);
    pushup(rt);
}

ll Quary(int L, int R, int l, int r, int rt) {
    if(L <= l && r <= R) return tree[rt].maxi;
    int mid = (l + r) >> 1; 
    ll a = -INF, b = -INF;
    if(L <= mid) a = Quary(L, R, l, mid, rt << 1);
    if(R > mid) b = Quary(L, R, mid + 1, r, rt << 1 | 1);
    return max(a, b);
}

int main() {
    scanf("%d %lld", n, D);
    build(1, N, 1);
    for(int i = 1; i <= n; i++) {
        char opt; ll x;
        cin >> opt; scanf("%lld", &x);
        if(opt == 'A') {
            x += t; x %= D; cnt++;
            upd(cnt, x, 1, N, 1);
        } else {
            t = Quary(cnt - x + 1, cnt, 1, N, 1);
            printf("%lld\n", t);
        }
    }
    return 0;
}