CF719E. Sasha and Array [线段树维护矩阵]

CF719E. Sasha and Array

题意：

对长度为 n 的数列进行 m 次操作, 操作为:

1. a[l..r] 每一项都加一个常数 C, 其中 0 ≤ C ≤ 10^9
2. 求 F[a[l]]+F[a[l+1]]+...F[a[r]] mod 1e9+7 的余数

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矩阵快速幂求斐波那契

矩阵满足乘法分配律和结合律！

所以可以每个节点维护矩阵/矩阵和，区间加相当于区间乘矩阵

注意：不要把快速幂写在里面，复杂度平添一个log。把\(B^C\)算出来之后传进去就好了

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long ll;
#define lc x<<1
#define rc x<<1|1
#define mid ((l+r)>>1)
#define lson lc, l, mid
#define rson rc, mid+1, r
const int N = 1e5+5, P = 1e9+7;
int n, m;
struct Matrix {
	ll a[2][2];
	ll* operator [](int x) {return a[x];}
	Matrix(int p=0) {
		if(!p) a[0][0] = a[0][1] = a[1][0] = a[1][1] = 0;
		else a[0][0] = a[1][1] = 1, a[0][1] = a[1][0] = 0;
	}
} B;
Matrix operator + (Matrix a, Matrix b) {
	Matrix c;
	for(int i=0; i<2; i++)
		for(int j=0; j<2; j++)
			c[i][j] = (a[i][j] + b[i][j]) %P;
	return c;
}
Matrix operator * (Matrix a, Matrix b) {
	Matrix c;
	for(int i=0; i<2; i++)
		for(int j=0; j<2; j++) {
			ll &x = c[i][j];
			for(int k=0; k<2; k++)
				x = (x + a[i][k] * b[k][j] %P) %P;
		}
	return c;
}
Matrix operator ^ (Matrix a, int b) {
	Matrix ans(1);
	ans[0][0] = ans[1][1] = 1;
	for(; b; b>>=1, a=a*a)
		if(b & 1) ans = ans*a;
	return ans;
}
struct meow {
	Matrix f;
	Matrix v;
	int c;
	meow() {f[0][0] = 1; v[0][0] = v[1][1] = 1;}
} t[N<<2];
void paint(int x, int l, int r, int d, Matrix &v) {
	t[x].c += d;
	t[x].v = v * t[x].v;
	t[x].f = v * t[x].f;
}
void push_down(int x, int l, int r) {
	if(t[x].c) {
		paint(lson, t[x].c, t[x].v);
		paint(rson, t[x].c, t[x].v);
		t[x].c = 0;
		t[x].v = Matrix(1);
	}
}
void merge(int x) {
	t[x].f = t[lc].f + t[rc].f;
}
void build(int x, int l, int r) {
	if(l == r) {
		cin >> t[x].c;
		if(t[x].c > 1) t[x].f = (B ^ (t[x].c - 1)) * t[x].f;
	} else {
		build(lson);
		build(rson);
		merge(x);
	}
}
void Add(int x, int l, int r, int ql, int qr, int d, Matrix &v) {
	if(ql <= l && r <= qr) paint(x, l, r, d, v);
	else {
		push_down(x, l, r);
		if(ql <= mid) Add(lson, ql, qr, d, v);
		if(mid < qr)  Add(rson, ql, qr, d,v );
		merge(x);
	}
}
ll Que(int x, int l, int r, int ql, int qr) {
	if(ql <= l && r <= qr) return t[x].f[0][0];
	else {
		push_down(x, l, r);
		ll ans = 0;
		if(ql <= mid) ans = (ans + Que(lson, ql, qr)) %P;
		if(mid < qr)  ans = (ans + Que(rson, ql, qr)) %P;
		return ans;
	}
}
int main() {
	//freopen("in", "r", stdin);
	ios::sync_with_stdio(false); cin.tie(); cout.tie();
	B[0][0] = B[0][1] = B[1][0] = 1;
	cin >> n >> m;
	build(1, 1, n);
	for(int i=1; i<=m; i++) {
		int tp, l, r, x;
		cin >> tp >> l >> r;
		if(tp == 1) {
			cin >> x;
			Matrix t = B^x;
			Add(1, 1, n, l, r, x, t);
		} else cout << Que(1, 1, n, l, r) << '\n';
	}