ZOJ-3772 Calculate the Function(线段树，矩阵乘法)

Calculate the Function

---

Time Limit: 2 Seconds       Memory Limit: 65536 KB

---

You are given a list of numbers A₁ A₂ .. A_N and M queries. For the i-th query:

• The query has two parameters L_i and R_i.
• The query will define a function F_i(x) on the domain [L_i, R_i] ∈ Z.
• F_i(L_i) = A_Li
• F_i(L_i + 1) = A_{(Li + 1)}
• for all x >= L_i + 2, F_i(x) = F_i(x - 1) + F_i(x - 2) × A_x

You task is to calculate F_i(R_i) for each query. Because the answer can be very large, you should output the remainder of the answer divided by 1000000007.

Input

There are multiple test cases. The first line of input is an integer T indicates the number of test cases. For each test case:

The first line contains two integers N, M (1 <= N, M <= 100000). The second line contains N integers A₁ A₂ .. A_N (1 <= A_i <= 1000000000).

The next M lines, each line is a query with two integer parameters L_i, R_i (1 <= L_i <= R_i <= N).

Output

For each test case, output the remainder of the answer divided by 1000000007.

Sample Input

1
4 7
1 2 3 4
1 1
1 2
1 3
1 4
2 4
3 4
4 4

Sample Output

1
2
5
13
11
4
4

题意：

给你n个数，A1,A2......An，然后有m次查询，每次查询给你一个L和R，然后令F(L)=AL,F(L+1)=AL+1。

然后告诉你你递推式F(n)=F(n-1)+F(n-2)*An，求F(R)。

思路：

看到这个数据范围，第一反应是矩阵快速幂，后来发现每次乘的矩阵都在变化，所以想到应该是用线段树去优化。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <iterator>
#include <map>
#include <algorithm>
#include <set>
#include <functional>
#include <time.h>
#define lson rt<<1
#define rson rt<<1|1
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int INF = 1e9 + 5;
const int MAXN = 100005;
const LL MOD = 1000000007;
const double eps = 1e-8;
const double PI = acos(-1.0);
LL gcd(LL a, LL b) { return b == 0 ? a : gcd(b, a%b); }
LL ppow(LL a, LL b) { LL res = 1; for (int i = 1; i <= b; i++)	res *= a; return res; }
LL quick_mod(LL a, LL b, LL c) { LL ans = 1; while (b) { if (b % 2 == 1)ans = (ans*a) % c; b /= 2; a = (a*a) % c; }return ans; }

struct Mat {
	LL n, m;   //行列  
	LL mat[3][3];
};
Mat operator *(Mat a, Mat b) {  //矩阵相乘    
	Mat c;
	memset(c.mat, 0, sizeof(c.mat));
	c.n = a.n, c.m = b.m;
	for (int i = 1; i <= a.n; i++) {
		for (int j = 1; j <= b.m; j++) {
			for (int k = 1; k <= a.m; k++) {
				c.mat[i][j] += (a.mat[i][k] * b.mat[k][j]) % MOD;
				c.mat[i][j] = (c.mat[i][j] % MOD + MOD) % MOD;
			}
		}
	}
	return c;
}

struct node
{
	int l, r, mid;
	Mat res;
}t[MAXN << 2];

Mat a[MAXN];
Mat ans;

void pushup(int rt)
{
	t[rt].res = t[rson].res*t[lson].res;
}
void build(int l, int r, int rt)
{
	int m = (l + r) >> 1;
	t[rt].l = l, t[rt].r = r;
	t[rt].mid = m;
	if (l == r)
	{
		t[rt].res = a[l];
		return;
	}
	build(l, m, lson);
	build(m + 1, r, rson);
	pushup(rt);
}
void query(int p, int q, int rt)
{
	if (p <= t[rt].l&&t[rt].r <= q)
	{
		ans = ans*t[rt].res;
		return;
	}
	if (q>t[rt].mid)
		query(p, q, rson);
	if (p <= t[rt].mid)
		query(p, q, lson);
}

int main()
{
	int T;
	scanf("%d", &T);
	int n, m;
	int A, B;
	Mat p;
	p.m = 1;
	p.n = 2;
	while (T--)
	{
		scanf("%d%d", &n, &m);
		for (int i = 1; i <= n; i++)
		{
			a[i].m = 2;
			a[i].n = 2;
			a[i].mat[1][1] = 1;
			scanf("%lld", &a[i].mat[1][2]);
			a[i].mat[2][1] = 1;
			a[i].mat[2][2] = 0;
		}
		build(1, n, 1);
		for (int i = 1; i <= m; i++)
		{
			scanf("%d%d", &A, &B);
			p.mat[1][1] = a[A + 1].mat[1][2];
			p.mat[2][1] = a[A].mat[1][2];
			if (B - A  == 0)
				printf("%lld\n", p.mat[2][1] % MOD);
			else if (B - A  == 1)
				printf("%lld\n", p.mat[1][1] % MOD);
			else
			{
				ans.m = 2;
				ans.n = 2;
				ans.mat[1][1] = 1;
				ans.mat[1][2] = 0;
				ans.mat[2][1] = 0;
				ans.mat[2][2] = 1;
				query(A + 2, B, 1);
				p = ans*p;
				printf("%lld\n", p.mat[1][1] % MOD);
			}
		}
	}
}