luogu P3168 [CQOI2015] 任务查询系统

传送门

区间修改单点查询的主席树

改个差分就行了

首先强制在线的是查询

所以修改可以一次全读进来然后离散并且插进去 没有影响

这里的话先全修改完再查询 可以放弃树状数组直接维护差分

然后主席树维护区间数字个数和整体和

最后分到叶子节点的时候注意去对应个数个数字加进去就行

然后有个操作就是一次把所有这个点上的差分数组都加进去

不开long long ***

Code:

#include<cstdio>
#include<iostream>
#include<cstring>
#include<vector>
#include<algorithm>
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
using namespace std;
typedef long long ll;
int read() {
    int as = 0,fu = 1;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') fu = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        as = as * 10 + c - '0';
        c = getchar();
    }
    return as * fu;
}
const int N = 1000005;
const int M = 16000005;
//head
int n,Q,T;
int ls[M],rs[M],sze[M],sum[M];
int tot,h[N],rt[N];
void ins(int l,int r,int &t,int pre,int q,int v) {
    t = ++tot,ls[t] = ls[pre],rs[t] = rs[pre];
    sze[t] = sze[pre] + v,sum[t] = sum[pre] + h[q] * v;
    if(l == r) return;
    int m = l+r >> 1;
    if(q <= m) ins(l,m,ls[t],ls[pre],q,v);
    else ins(m+1,r,rs[t],rs[pre],q,v);
}

int query(int l,int r,int t,int k) {
    if(k <= 0) return 0;
    if(k >= sze[t]) return sum[t];
    if(l == r) return sum[t] * k / sze[t];
    int m = l+r >> 1,ans = query(l,m,ls[t],k);
    if(k > sze[ls[t]]) ans += query(m+1,r,rs[t],k - sze[ls[t]]);
    return ans;
}
vector<int> a[N];
int main() {
    // freopen("in.in","r",stdin);
    // freopen("a.out","w",stdout);
    Q = read(),n = read();
    rep(i,1,Q) {
        int x = read(),y = read(),z = read();
        a[x].push_back(z),a[y+1].push_back(-z);
        h[i] = z;
    }
    sort(h+1,h+Q+1),T = unique(h+1,h+Q+1) - (h+1);
    rep(i,1,n) {
        rt[i] = rt[i-1];
        rep(j,0,(int)a[i].size() - 1) {
            int k = (a[i][j] < 0 ? -1 : 1);
            a[i][j] = max(a[i][j],-a[i][j]);
            ins(1,T,rt[i],rt[i],lower_bound(h+1,h+T+1,a[i][j]) - h,k);
        }
    }
    int ans = 1;
    rep(i,1,n) {
        ll x = read(),a = read(),b = read(),c = read();
        int k = (a * ans + b) % c + 1;
        printf("%d
",ans = query(1,T,rt[x],k));
    }
    return 0;
}