一道好题
Solution
区间操作，容易想到线段树
但是普通的线段树好像无法维护这种区间操作
考虑将操作转换成矩阵乘法，再用线段树维护矩阵
这里的操作有两种转换方法：
1.转换成矩阵的乘法和加法：
线段树维护区间矩阵和
优势：空间和时间复杂度的常数要小
缺点：有优先级问题（类似于线段树模板1和线段树模板2的区别）
2.全部转换成矩阵的乘法：
线段树维护区间矩阵乘积
优势：只用维护区间乘积，无优先级问题
缺点：时空复杂度的常数比方法1略大
两种方法各有利弊，根据题目要求具体选择
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2.5e5 + 10;
const int MOD = 998244353;
typedef long long ll;
#define rint register int
int n, m;
#define lson (node << 1)
#define rson (node << 1 | 1)
#define mid ((l + r) >> 1)
inline int Add(int a, int b) { return (a += b) >= MOD ? a - MOD : a; }
struct Matrix {
int a[3][3], n, m;
inline void Clear(void) { memset(a, 0, sizeof(a)); }
inline Matrix(const int _n = 0, const int _m = 0) {
n = _n;
m = _m;
Clear();
}
inline void IOne(void) {
for (register int i = 0; i < n; ++i) a[i][i] = 1;
}
inline void Init(const int _n, const int _m) {
n = _n;
m = _m;
}
Matrix operator+(const Matrix &p) const {
Matrix c(n, m);
for (register int i = 0; i < n; ++i) {
for (register int j = 0; j < m; ++j) {
c.a[i][j] = Add(a[i][j], p.a[i][j]);
}
}
return c;
}
Matrix operator*(const int p) const {
Matrix c(n, m);
for (register int i = 0; i < n; ++i) {
for (register int j = 0; j < m; ++j) {
c.a[i][j] = (long long)a[i][j] * p % MOD;
}
}
return c;
}
Matrix operator*(const Matrix &p) const {
Matrix c(n, p.m);
for (register int i = 0; i < n; ++i) {
for (register int k = 0; k < m; ++k) {
for (register int j = 0; j < p.m; ++j)
c.a[i][j] = Add(c.a[i][j], (ll)a[i][k] * p.a[k][j] % MOD);
}
}
return c;
}
};
Matrix data[MAXN << 2], tmul[MAXN << 2], tadd[MAXN << 2];
void build(int node, int l, int r) {
data[node].Init(1, 3);
tmul[node].Init(3, 3);
tmul[node].IOne();
tadd[node].Init(1, 3);
if (l == r) {
for (register int i = 0; i < 3; ++i) scanf("%d", &data[node].a[0][i]);
return;
}
build(lson, l, mid);
build(rson, mid + 1, r);
data[node] = data[lson] + data[rson];
}
inline void pushdown(int node, int l, int r) {
tmul[lson] = tmul[lson] * tmul[node];
tmul[rson] = tmul[rson] * tmul[node];
tadd[lson] = tadd[lson]* tmul[node] + tadd[node];
tadd[rson] = tadd[rson] * tmul[node] + tadd[node];
data[lson] = data[lson] * tmul[node] + tadd[node] * (mid - l + 1);
data[rson] = data[rson] * tmul[node] + tadd[node] * (r - mid);
tmul[node].Clear();
tmul[node].IOne();
tadd[node].Clear();
}
void change(int node, int l, int r, int from, int to, Matrix val, int type) {
if (from > r || to < l)
return;
if (from <= l && r <= to) {
if (type == 0)
tmul[node] = tmul[node] * val, tadd[node] = tadd[node] * val, data[node] = data[node] * val;
else
tadd[node] = tadd[node] + val, data[node] = data[node] + val * (r - l + 1);
return;
}
pushdown(node, l, r);
change(lson, l, mid, from, to, val, type);
change(rson, mid + 1, r, from, to, val, type);
data[node] = data[lson] + data[rson];
}
Matrix query(int node, int l, int r, int from, int to) {
if (from <= l && r <= to)
return data[node];
pushdown(node, l, r);
Matrix ret(1, 3);
if (from <= mid)
ret = ret + query(lson, l, mid, from, to);
if (to > mid)
ret = ret + query(rson, mid + 1, r, from, to);
return ret;
}
inline void init(void) {
scanf("%d", &n);
build(1, 1, n);
scanf("%d", &m);
}
inline void work(void) {
Matrix m1(3, 3), m2(3, 3), m3(3, 3), p4(1, 3), m5(3, 3), m6(3, 3), p6(1, 3);
m1.a[0][0] = m1.a[1][0] = m1.a[1][1] = m1.a[2][2] = 1;
m2.a[0][0] = m2.a[1][1] = m2.a[2][1] = m2.a[2][2] = 1;
m3.a[0][0] = m3.a[1][1] = m3.a[0][2] = m3.a[2][2] = 1;
m5.a[0][0] = m5.a[2][2] = m6.a[0][0] = m6.a[1][1] = 1;
int op, l, r;
for (register int i = 1; i <= m; ++i) {
scanf("%d%d%d", &op, &l, &r);
if (op == 1)
change(1, 1, n, l, r, m1, 0);
if (op == 2)
change(1, 1, n, l, r, m2, 0);
if (op == 3)
change(1, 1, n, l, r, m3, 0);
if (op == 4)
scanf("%d", &p4.a[0][0]), change(1, 1, n, l, r, p4, 1);
if (op == 5)
scanf("%d", &m5.a[1][1]), change(1, 1, n, l, r, m5, 0);
if (op == 6)
scanf("%d", &p6.a[0][2]), change(1, 1, n, l, r, m6, 0), change(1, 1, n, l, r, p6, 1);
if (op == 7) {
Matrix now = query(1, 1, n, l, r);
printf("%d %d %d\n", now.a[0][0], now.a[0][1], now.a[0][2]);
}
}
}
int main() {
init();
work();
return 0;
}