codeforces 718c 矩阵快速幂套线段树

题意 给出一个序列ai，定义斐波那契数列，求sigma f(a_i)
还有可能令一段a_i加上一个数字。
斐波那契数列可以有矩阵快速幂加速求出，复杂度logn，我们这里定义mat为加速矩阵，那么对一段求和便可以写成sigma mat_i，对于一段区间加上一个数字，实际上等同于给每一个mat在乘上一个矩阵。
实际上变成维护区间加和的题目，只不过每一个节点储存一个矩阵。

    #include<iostream>
    #include<cstring>
    #include<algorithm>
    #include<cmath>
    #include<cstdlib>
    #include<climits>
    #include<stack>
    #include<vector>
    #include<queue>
    #include<set>
    #include<map>
    //#include<regex>
    #include<cstdio>
    #define up(i,a,b)  for(int i=a;i<b;i++)
    #define dw(i,a,b)  for(int i=a;i>b;i--)
    #define upd(i,a,b) for(int i=a;i<=b;i++)
    #define dwd(i,a,b) for(int i=a;i>=b;i--)
    //#define local
    typedef long long ll;
    const double esp = 1e-6;
    const double pi = acos(-1.0);
    const int INF = 0x3f3f3f3f;
    const int inf = 1e9;
    using namespace std;
    ll read()
    {
    	char ch = getchar(); ll x = 0, f = 1;
    	while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
    	while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
    	return x * f;
    }
    typedef pair<int, int> pir;
    #define lson l,mid,root<<1
    #define rson mid+1,r,root<<1|1
    #define lrt root<<1
    #define rrt root<<1|1
    const int N = 1e5 + 10;
    const int mod = 1e9 + 7;
    struct matrix {
    	ll a[2][2];
    	void init() { memset(a, 0, sizeof(a)); }
    	void initfib() { a[0][0] = a[0][1] = a[1][0] = 1; a[1][1] = 0; }
    	void one() { a[0][0] = a[1][1] = 1; a[0][1] = a[1][0] = 0; }
    	matrix operator +(const matrix &m)const {
    		matrix b; b.init();
    		up(i, 0, 2)up(j, 0, 2)b.a[i][j] = (a[i][j] + m.a[i][j]) % mod; return b;
    	}
    	matrix operator *(const matrix &m)const {
    		matrix b; b.init();
    		up(i, 0, 2)up(j, 0, 2)up(k, 0, 2)b.a[i][j] = (b.a[i][j] + a[i][k]* m.a[k][j] % mod) % mod;return b;
    	}
    	matrix operator ^(const ll &k)const
    	{
    		ll num = k;
    		matrix res, now; res.one(); now = *this;
    		while (num)
    		{
    			if (num & 1)res = res * now;
    			now = now * now;
    			num >>= 1;
    		}
    		return res;
    	}
    }tree[N<<2],lazy[N<<2];
    matrix isone;
    bool same(matrix a, matrix b) { up(i, 0, 2)up(j, 0, 2)if (a.a[i][j] != b.a[i][j])return false; return true; }
    void pushup(int root)
    {
    	tree[root] = tree[lrt] + tree[rrt];
    }
    void pushdown(int root)
    {
    	if (!same(isone, lazy[root]))
    	{
    		lazy[lrt] = lazy[lrt] * lazy[root];
    		lazy[rrt] = lazy[rrt] * lazy[root];
    		tree[lrt] = tree[lrt] * lazy[root];
    		tree[rrt] = tree[rrt] * lazy[root];
    		lazy[root] = isone;
    	}
    }
    void build(int l, int r, int root)
    {
    	lazy[root].one();
    	if (l == r)
    	{
    		tree[root].one(); return;
    	}
    	int mid = (l + r) >> 1;
    	build(lson);
    	build(rson);
    	pushup(root);
    }
    void update(int l,int r,int root,int lf,int rt,matrix x)
    {
    	if (lf <= l && r <= rt)
    	{
    		tree[root] = tree[root] * x;
    		lazy[root] = lazy[root] * x;
    		return;
    	}
    	pushdown(root);
    	int mid = (l + r) >> 1;
    	if (lf <= mid)update(lson, lf, rt, x);
    	if (rt > mid)update(rson, lf, rt, x);
    	pushup(root);
    }
    matrix querry(int l, int r, int root,int lf,int rt)
    {
    	if (lf <= l && r <= rt)
    	{
    		return tree[root];
    	}
    	pushdown(root);
    	int mid=(l + r) >> 1;
    	matrix ans; ans.init();
    	if (lf <= mid)ans = ans + querry(lson, lf, rt);
    	if (rt > mid)ans = ans + querry(rson, lf, rt);
    	return ans;
    }
    int n, m;
    ll a[N]; 
    int main()
    {
    	n = read(), m = read(); isone.one();
    	ll tp, l, r,x;
    	build(1, n, 1);
    	up(i, 0, n)a[i] = read();
    	matrix tmp; tmp.initfib();
    	up(i, 0, n)tmp = tmp ^ a[i], update(1,n, 1,i + 1, i + 1, tmp), tmp.initfib();
    	while (m--)
    	{
    		tp = read();
    		if (tp == 1) {
    			l = read(), r = read(), x = read(); matrix tmp; tmp.initfib(); tmp = tmp ^ x; update(1, n, 1, l, r, tmp);
    		}
    		else
    		{
    			l = read(), r = read(); printf("%lld\n", querry(1, n, 1, l,r).a[1][0]);
    		}
    	}
    	return 0;
    }