bzoj 3821 玄学 [线段树+归并排序]

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题意：给出一个大小为n的数组，有两种操作：1.对L到R区间内的数全部进行(a*a[i]+b)%m的操作 2.询问区间L到R第k个数的大小

题解：可以对每次操作建立关于时间的线段树，每个节点【L，R】保存第L次操作到第R次操作对数组的影响（用段来表示，同一段内a和b的值相同，设T为当前时间段的操作次数，则最多2*T个段），用归并排序的思想对线段树左子树与右子树进行合并，每次插入一个操作时需要进行logn次归并，每次归并平均 logn的时间所以总的时间复杂度N*logn*logn，查找的时间为logn查找节点logn查找第k个位置，空间复杂度n*logn。

#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[N], 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<<1], t2 = L[x<<1|1], r1 = R[x<<1], r2 = R[x<<1|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;  }

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