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; 
}