【bzoj 3821】玄学 线段树

给一个数列ai。要求维护（在线）：

1.将ai到aj的值变为(ax + b) mod m

2.问执行操作i到操作j后ak的权值

对操作建线段树，一个线段树节点只在加入操作r后update，一个节点可以用一个长度为O(l)的数组维护，询问时二分下即可。log平方n。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#define Rep(i, x, y) for (int i = x; i <= y; i ++)
#define Dwn(i, x, y) for (int i = x; i >= y; i --)
#define RepE(i, x) for(int i = pos[x]; i; i = g[i].nex)
using namespace std;
typedef long long LL;
const int N = 600005, M = N * 25;
int n, m, T0, L[N*4], R[N*4], dl[M], dr[M], ql, qr, qa, qb, dz, mod, ty, mx, a
, q, T, lans; LL da[M], db[M], a1, b1;
int read() {
int x = 0; char c; for (c = getchar(); c < '0' || c > '9'; c = getchar()) ;
for (; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';
return x;
}
void Upd(int x) {
int t1 = L[x + x], t2 = L[x + x + 1], r1 = R[x + x], r2 = R[x + x + 1], l0 = 1;
L[x] = dz + 1;
while (t1 <= r1 || t2 <= r2) {
dl[++ dz] = l0, dr[dz] = min(dr[t1], dr[t2]);
da[dz] = da[t1] * da[t2] % mod;
db[dz] = (da[t2] * db[t1] + db[t2]) % mod;
l0 = dr[dz] + 1;
if (t2 > r2 || (t1 <= r1 && dr[t1] < dr[t2])) t1 ++;
else if (t1 > r1 || dr[t2] < dr[t1]) t2 ++;
else t1 ++, t2 ++;
} R[x] = dz;
}
void Modify(int x, int l, int r) {
if (l == r) {
L[x] = dz + 1;
if (ql > 1) dl[++ dz] = 1, dr[dz] = ql - 1, da[dz] = 1;
dl[ ++ dz ] = ql, dr[dz] = qr, da[dz] = qa, db[dz] = qb;
if (qr < n) dl[++ dz] = qr + 1, dr[dz] = n, da[dz] = 1;
R[x] = dz;
return ;
}
int mid = l + r >> 1;
if (q <= mid) Modify(x+x, l, mid); else Modify(x+x+1, mid+1, r);
if (q == r) Upd(x);
}
int Find(int l, int r) {
while (l < r) {
int mid = l + r >> 1;
if (dr[mid] < qa) l = mid + 1; else r = mid;
} return l;
}
void Qry(int x, int l, int r) {
if (ql > r || qr < l) return ;
if (ql <= l && r <= qr) {
int k = Find(L[x], R[x]);
a1 = a1 * da[k] % mod, b1 = (b1 * da[k] + db[k]) % mod;
return ;
}
int mid = l + r >> 1;
Qry(x+x, l, mid), Qry(x+x+1, mid+1, r);
}
int main()
{
scanf ("%d%d%d", &T0, &n, &mod);
Rep(i, 1, n) scanf ("%d", &a[i]);
scanf ("%d", &T); mx = T;
while (T --) {
ty = read(), ql = read(), qr = read(), qa = read();
if (T0 & 1) ql ^= lans, qr ^= lans;
if (ty == 1) {
qb = read(), q ++; Modify(1, 1, mx);
} else {
if (T0 & 1) qa ^= lans; a1 = 1, b1 = 0, Qry(1, 1, mx);
lans = (a1 * a[qa] + b1) % mod; printf("%d\n", lans);
}
}

return 0;
}