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區間修改單點查詢的主席樹

改個差分就行了

首先強制線上的是查詢

所以修改可以一次全讀進來然後離散並且插進去 沒有影響

這裡的話先全修改完再查詢 可以放棄樹狀陣列直接維護差分

然後主席樹維護區間數字個數和整體和

最後分到葉子節點的時候注意去對應個數個數字加進去就行

然後有個操作就是一次把所有這個點上的差分陣列都加進去

不開long long ***

Code:

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<vector>
 5
 
 #include<algorithm>
 6 #define rep(i,a,n) for(int i = a;i <= n;i++)
 7 #define per(i,n,a) for(int i = n;i >= a;i--)
 8 using namespace std;
 9 typedef long long ll;
10 int read() {
11     int as = 0,fu = 1;
12     char c = getchar();
13     while(c < '0' || c > '9') {
14         if(c == '
 
-') fu = -1;
15         c = getchar();
16     }
17     while(c >= '0' && c <= '9') {
18         as = as * 10 + c - '0';
19         c = getchar();
20     }
21     return as * fu;
22 }
23 const int N = 1000005;
24 const int M = 16000005;
25 //head
26 int n,Q,T;
27 int ls[M],rs[M],sze[M],sum[M];
28 int tot,h[N],rt[N];
 
29 void ins(int l,int r,int &t,int pre,int q,int v) {
30     t = ++tot,ls[t] = ls[pre],rs[t] = rs[pre];
31     sze[t] = sze[pre] + v,sum[t] = sum[pre] + h[q] * v;
32     if(l == r) return;
33     int m = l+r >> 1;
34     if(q <= m) ins(l,m,ls[t],ls[pre],q,v);
35     else ins(m+1,r,rs[t],rs[pre],q,v);
36 }
37 
38 int query(int l,int r,int t,int k) {
39     if(k <= 0) return 0;
40     if(k >= sze[t]) return sum[t];
41     if(l == r) return sum[t] * k / sze[t];
42     int m = l+r >> 1,ans = query(l,m,ls[t],k);
43     if(k > sze[ls[t]]) ans += query(m+1,r,rs[t],k - sze[ls[t]]);
44     return ans;
45 }
46 vector<int> a[N];
47 int main() {
48     // freopen("in.in","r",stdin);
49     // freopen("a.out","w",stdout);
50     Q = read(),n = read();
51     rep(i,1,Q) {
52         int x = read(),y = read(),z = read();
53         a[x].push_back(z),a[y+1].push_back(-z);
54         h[i] = z;
55     }
56     sort(h+1,h+Q+1),T = unique(h+1,h+Q+1) - (h+1);
57     rep(i,1,n) {
58         rt[i] = rt[i-1];
59         rep(j,0,(int)a[i].size() - 1) {
60             int k = (a[i][j] < 0 ? -1 : 1);
61             a[i][j] = max(a[i][j],-a[i][j]);
62             ins(1,T,rt[i],rt[i],lower_bound(h+1,h+T+1,a[i][j]) - h,k);
63         }
64     }
65     int ans = 1;
66     rep(i,1,n) {
67         ll x = read(),a = read(),b = read(),c = read();
68         int k = (a * ans + b) % c + 1;
69         printf("%d\n",ans = query(1,T,rt[x],k));
70     }
71     return 0;
72 }