THUSC 2017 大魔法师

一个序列，每个物品有三个权值 $A,B,C$

要求维护：

1.区间 $A_i+=B_i$

2.区间 $B_i+=C_i$

3.区间 $C_i+=A_i$

4.区间 $A_i+=v$

5.区间 $B_i \times = v$

6.区间 $C_i = v$

7.询问区间 $A,B,C$ 各自的和

线段树，每个点维护 $A,B,C,区间长度$

每次修改相当于区间乘一个转移矩阵

时间复杂度 $O(16nlogn)$

垫底

#include <bits/stdc++.h>
#define LL long long
using namespace std;
#define rep(i, s, t) for (register int i = (s), i##end = (t); i <= i##end; ++i)
#define dwn(i, s, t) for (register int i = (s), i##end = (t); i >= i##end; --i)
inline int read() {
    int x = , f = ; char ch = getchar();
    for (; !isdigit(ch); ch = getchar())if (ch == '-')f = -f;
    for (; isdigit(ch); ch = getchar()) x =  * x + ch - '';
    return x * f;
}
const int mod = , maxn = 2.5e5 + ;
#define ls (x << 1)
#define rs ((x << 1) | 1)
int n, q, A[maxn], B[maxn], C[maxn];
struct Matrix {
    int a[][];
    Matrix() {memset(a, , sizeof(a));}
    Matrix operator * (const Matrix &b) const {
        Matrix c;
        rep(i, , ) rep(j, , ) rep(k, , )
            (c.a[i][j] += (1LL * a[i][k] * b.a[k][j] % mod)) %= mod;
        return c;
    }
    Matrix operator + (const Matrix &b) const {
        Matrix c;
        rep(i, , ) rep(j, , ) c.a[i][j] = (a[i][j] + b.a[i][j]) % mod;
        return c;
    }
}tag[maxn << ];
int seg[maxn << ][];
void mul(int *f, Matrix gg) {
    int tmp[] = {, , , };
    rep(i, , ) rep(j, , ) (tmp[j] += (1LL * f[i] * gg.a[i][j] % mod)) %= mod;
    rep(i, , ) f[i] = tmp[i];
}
inline int clear(Matrix x) {
    if (!(x.a[][] ==  && x.a[][] ==  && x.a[][] ==  && x.a[][] == ))return false;
    if (x.a[][] || x.a[][] || x.a[][] || x.a[][] || x.a[][] || x.a[][])return false;
    if (x.a[][] || x.a[][] || x.a[][] || x.a[][] || x.a[][] || x.a[][]) return false;
    return true;
}
inline void pushup(int x) {
    rep(i, , ) seg[x][i] = (seg[ls][i] + seg[rs][i]) % mod;
}
inline void pushdown(int x) {
    if(clear(tag[x])) return;
    tag[ls] = tag[ls] * tag[x], tag[rs] = tag[rs] * tag[x];
    mul(seg[ls], tag[x]), mul(seg[rs], tag[x]);
    rep(i, , ) rep(j, , ) tag[x].a[i][j] = (i == j);
}
inline void build(int x, int l, int r) {
    if(l == r) {
        seg[x][] = A[l]; seg[x][] = B[l]; seg[x][] = C[l]; seg[x][] = ;
        return;
    }
    int mid = (l + r) >> ;
    build(ls, l, mid); build(rs, mid + , r);
    pushup(x);
}
Matrix cur; int res[];
inline void update(int x, int l, int r, int L, int R) {
    if(L <= l && r <= R) {
        mul(seg[x], cur);
        tag[x] = tag[x] * cur;
        return;
    }
    pushdown(x);
    int mid = (l + r) >> ;
    if(L <= mid) update(ls, l, mid, L, R);
    if(R > mid) update(rs, mid + , r, L, R);
    pushup(x);
}
inline void query(int x, int l, int r, int L, int R) {
    if(L <= l && r <= R) {
        rep(i, , ) (res[i] += seg[x][i]) %= mod;
        return;
    }
    pushdown(x);
    int mid = (l + r) >> ;
    if(L <= mid) query(ls, l, mid, L, R);
    if(R > mid) query(rs, mid + , r, L, R);
}
int main() {
    n = read();
    rep(i, , (n<<)) rep(j, , ) rep(k, , ) tag[i].a[j][k] = (j == k);
    rep(i, , n) A[i] = read(), B[i] = read(), C[i] = read();
    build(, , n);
    q = read();
    while(q--) {
        int opt = read(), l = read(), r = read();
        rep(i, , ) rep(j, , ) cur.a[i][j] = (i == j);
        if(opt == ) cur.a[][]++;
        else if(opt == ) cur.a[][]++;
        else if(opt == ) cur.a[][]++;
        else if(opt == ) (cur.a[][] += read()) %= mod;
        else if(opt == ) cur.a[][] = read();
        else if(opt == ) cur.a[][] = , cur.a[][] = read();
        if(opt != ) update(, , n, l, r);
        if(opt == ) {
            rep(i, , ) res[i] = ;
            query(, , n, l, r);
            printf("%d %d %d\n", res[], res[], res[]);
            continue;
        }
    }
}