已經菜爆了只會做簡單題…

Description

每個位置有三種屬性, 要支持把一個屬性加到另一個上, 或者屬性加, 屬性乘, 屬性賦值, 都是區間操作, 還有查詢區間和.

Solution

維護一個$4 \times 4$的矩陣, 分別是三種屬性和外加的單位, 第4項在單點設置爲1就好了.

然後看情況轉移一下.

順便這題卡常, 實現要注意細節.

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#pragma GCC optimize(3)
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#define mod 998244353

using namespace std;
typedef long long ll;

const int maxn = 25e+4 + 5;

inline void add(int &x, int y) {
  x += y;
  if (x >= mod)
    x -= mod;
}

struct matrix {
  int a[4][4];
  matrix() {
  }
  int* operator[](int x) {
    return a[x];
  }
  inline void empty() {
    memset(a, 0, sizeof a);
  }
  inline matrix operator*(matrix rhs) {
    matrix ret;
    ret.empty();
    for (int k = 0; k < 4; ++k)
      for (int i = 0; i < 4; ++i)
        for (int j = 0; j < 4; ++j)
          add(ret[i][j], 1ll * a[i][k] * rhs[k][j] % mod);
    return ret;
  }
  inline matrix operator*=(matrix rhs) {
    matrix ret;
    ret.empty();
    for (int i = 0; i < 4; ++i)
      for (int k = 0; k < 4; ++k)
        if (a[i][k])
          for (int j = 0; j < 4; ++j)
            add(ret[i][j], 1ll * a[i][k] * rhs[k][j] % mod);
    return (*this) = ret;   
  }
  inline bool E() {
    /*for (int i = 0; i < 4; ++i)
      for (int j = 0; j < 4; ++j) {
        if (i == j && a[i][j] != 1) return false;
        else if (i != j && a[i][j]) return false;
      }*/
    if (! (a[0][0] == 1 && a[1][1] == 1 && a[2][2] == 1 && a[3][3] == 1))
      return false;
    if (a[0][1] || a[0][2] || a[0][3] || a[1][0] || a[1][2] || a[1][3] ||
        a[2][0] || a[2][1] || a[2][3] || a[3][0] || a[3][1] || a[3][2])
      return false;
    return true;
  }
  inline void e() {
    memset(a, 0, sizeof a);
    a[0][0] = a[1][1] = a[2][2] = a[3][3] = 1;
  }
};
matrix trans, tag[maxn << 2];
int n, m, sum[maxn << 2][4];
int v[maxn][3], tmp[4], res[4];

inline int rd() {
  register int x = 0, c = getchar();
  while (!isdigit(c)) c = getchar();
  while (isdigit(c)) x = x * 10 + (c ^ 48), c = getchar();
  return x;
}
inline void puttag(int x, matrix t) {
  memset(tmp, 0, sizeof tmp);
  for (int i = 0; i < 4; ++i)
    for (int j = 0; j < 4; ++j)
      add(tmp[j], 1ll * sum[x][i] * t[i][j] % mod);
  //for (int i = 0; i < 4; ++i)
  //  sum[x][i] = tmp[i];  
  sum[x][0] = tmp[0];
  sum[x][1] = tmp[1];
  sum[x][2] = tmp[2];
  sum[x][3] = tmp[3];
}
inline void tagon(int cur, matrix t) {
  puttag(cur, t);
  tag[cur] *= t;
}
inline void pushdown(int cur) {
  if (tag[cur].E()) return;
  tagon(cur << 1, tag[cur]);
  tagon(cur << 1|1, tag[cur]);
  tag[cur].e();
}
inline void pushup(int cur) {
  sum[cur][0] = (sum[cur << 1][0] + sum[cur << 1|1][0]) % mod;
  sum[cur][1] = (sum[cur << 1][1] + sum[cur << 1|1][1]) % mod;
  sum[cur][2] = (sum[cur << 1][2] + sum[cur << 1|1][2]) % mod;
  sum[cur][3] = (sum[cur << 1][3] + sum[cur << 1|1][3]) % mod;
}
void build(int cur, int l, int r) {
  tag[cur].e();
  if (l == r) {
    sum[cur][0] = v[l][0];
    sum[cur][1] = v[l][1];
    sum[cur][2] = v[l][2];
    sum[cur][3] = 1;
    return;
  }
  int mid = (l + r) >> 1;
  build(cur << 1, l, mid);
  build(cur << 1|1, mid + 1, r);
  pushup(cur);
}
void update(int cur, int l, int r, int L, int R, matrix t) {
  if (L <= l && r <= R) {
    tagon(cur, t);
    return;
  }
  int mid = (l + r) >> 1; pushdown(cur);
  if (L <= mid) update(cur << 1, l, mid, L, R, t);
  if (R > mid) update(cur << 1|1, mid + 1, r, L, R, t);
  pushup(cur);
}
void query(int cur, int l, int r, int L, int R) {
  if (L <= l && r <= R) {
    add(res[0], sum[cur][0]);
    add(res[1], sum[cur][1]);
    add(res[2], sum[cur][2]);
    return;
  }
  int mid = (l + r) >> 1; pushdown(cur);
  if (L <= mid) query(cur << 1, l, mid, L, R);
  if (R > mid) query(cur << 1|1, mid + 1, r, L, R);
}

int main() {
  n = rd();
  for (int i = 1; i <= n; ++i)
    v[i][0] = rd(),
    v[i][1] = rd(),
    v[i][2] = rd();
  build(1, 1, n);
  m = rd();
  int opt, l, r, x;
  while (m--) {
    opt = rd(); l = rd(); r = rd();
    trans.e();
    if (opt == 1) {
      trans[1][0]++;
    } else if (opt == 2) {
      trans[2][1]++;
    } else if (opt == 3) {
      trans[0][2]++;
    } else if (opt == 4) {
      x = rd();
      trans[3][0] = (trans[3][0] + x) % mod;
    } else if (opt == 5) {
      x = rd();
      trans[1][1] = x;
    } else if (opt == 6) {
      x = rd();
      trans[2][2] = 0; trans[3][2] = x;
    }

    if (opt == 7) {
      memset(res, 0, sizeof res);
      query(1, 1, n, l, r);
      printf("%d %d %d\n", res[0], res[1], res[2]);
    } else {
      update(1, 1, n, l, r, trans);
    }
  } 
  return 0;
}