CF718C Sasha and Array 线段树 + 矩阵乘法

有两个操作：

• 将 $[l,r]$所有数 + $x$
• 求 $\sum_{i=l}^{r}fib(i)$

$n=m=10^5$  

直接求不好求，改成矩阵乘法的形式： 
$a_{i}=M^x\times fib_{1}$
直接用线段树维护 $M^x$ 即可.
因为矩阵乘法是满足结合律的： $A*B+A*C=A*(B+C)$

#include <cstdio>
#include <algorithm>
#include <cstring>
#define lson (now << 1)
#define rson (now << 1 | 1)
#define ll long long
#define setIO(s) freopen(s".in", "r" , stdin)
using namespace std;
const int N = 200003;
const ll mod = 1000000007;
struct Matrix
{
    int n, m;
    ll a[4][4];
    ll * operator[] (int x)  { return a[x]; }
    inline void re()
    {
        for(int i = 0; i < 3 ; ++i)
            for(int j = 0; j < 3; ++j) a[i][j] = 0;
    }
    inline void I()
    {
        re();
        for(int i = 0; i < 3 ; ++i) a[i][i] = 1ll;
    }
    Matrix friend operator * (Matrix a, Matrix b)
    {
        Matrix c;
        c.re();
        int i , j , k;
        for(i = 0; i < a.n ; ++i)
        {
            for(j = 0; j < b.m ; ++j)
                for(k = 0; k < a.m ; ++k)
                {
                    c[i][j] = (c[i][j] + (a[i][k] * b[k][j]) % mod) % mod;
                }
        }
        c.n = a.n , c.m = b.m;
        return c;
    }
    Matrix friend operator + (Matrix a, Matrix b)
    {
        Matrix c;
        c.n = 2, c.m = 1;
        for(int i = 0; i < c.n ; ++i)
        {
            for(int j = 0; j < c.m; ++j) c[i][j] = (a[i][j] + b[i][j]) % mod;
        }
        return c;
    }
}A[N], M, fib1, v;
Matrix operator ^ (Matrix a, int k)
{
    Matrix tmp;
    tmp.n = tmp.m = 2;
    for(tmp.I(); k ; a = a * a, k >>= 1) if(k & 1) tmp = tmp * a;
    return tmp;
}
inline void init()
{
    M.re(), fib1.re();
    M.n = M.m = 2;
    M[0][0] = 0, M[0][1] = 1, M[1][0] = 1, M[1][1] = 1;
    fib1.n = 2, fib1.m = 1;
    fib1[0][0] = 0, fib1[1][0] = 1;
}
int n , m ;
struct Node
{
    Matrix sum, lazy;
    int tag;
}t[N << 2];
inline void pushup(int l, int r, int now)
{
    int mid = (l + r) >> 1;
    t[now].sum = t[now << 1].sum;
    if(r > mid) t[now].sum = t[now].sum + t[now << 1 | 1].sum;
}
inline void mark(int now, Matrix f)
{
    t[now].sum = f * t[now].sum ;
    t[now].lazy = t[now].lazy * f;
    t[now].tag = 1;
}
inline void pushdown(int l, int r, int now)
{
    int mid = (l + r) >> 1;
    if(t[now].tag)
    {
        mark(lson, t[now].lazy);
        if(r > mid) mark(rson, t[now].lazy);
        t[now].lazy.I(), t[now].tag = 0;
    }
}
void build(int l, int r, int now)
{
    t[now].lazy.n = t[now].lazy.m = 2;
    t[now].lazy.I();
    t[now].tag = 0;
    if(l == r)
    {
        t[now].sum = A[l];
        return ;
    }
    int mid = (l + r) >> 1;
    if(mid >= l) build(l, mid, lson);
    if(r > mid) build(mid + 1, r, rson);
    pushup(l, r, now);
}
void update(int l, int r, int now, int L, int R)
{
    if(l >= L && r <= R)
    {
        mark(now , v);
        return ;
    }
    pushdown(l, r, now);
    int mid = (l + r) >> 1;
    if(L <= mid) update(l, mid, lson, L, R);
    if(R > mid) update(mid + 1, r, rson, L, R);
    pushup(l, r, now);
}
ll query(int l, int r, int now, int L, int R)
{
    if(l >= L && r <= R)  return t[now].sum[1][0];
    pushdown(l, r, now);
    int mid = (l + r) >> 1;
    ll g = 0;
    if(L <= mid) g += query(l, mid, lson, L, R);
    if(R  > mid) g += query(mid + 1, r, rson, L, R);
    return (g % mod);
}
int main()
{
    // setIO("input");
    init();
    int i , j, x;
    scanf("%d%d", &n, &m);
    for(i = 1; i <= n ; ++i)
    {
        scanf("%d", &x), A[i] = (M ^ (x - 1)) * fib1;
    }
    build(1, n, 1);
    for(int cas = 1, op , l, r, x; cas <= m ; ++cas)
    {
        scanf("%d", &op);
        if(op == 1)
        {
            scanf("%d%d%d", &l, &r, &x);
            if(x == 0) continue;
            v = M ^ x;
            update(1, n, 1, l, r);
        }
        if(op == 2)
        {
            scanf("%d%d", &l, &r);
            printf("%lld\n", query(1, n, 1, l, r));
        }
    }
    return 0;
}