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

Calculate the FunctionTime Limit: 2 Seconds      Memory Limit: 65536 KBYou are given a list of numbers A1 A2 .. AN and M queries. For the i-th query:The query has two parameters Li and Ri.The query will define a function Fi(x) on the domain [Li, Ri] ∈ Z.Fi(Li) = ALiFi(Li + 1) = A(Li + 1)for all x >= Li + 2, Fi(x) = Fi(x - 1) + Fi(x - 2) × AxYou task is to calculate Fi(Ri) 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 .. AN (1<= Ai <= 1000000000).The next M lines, each line is a query with two integer parameters Li, Ri (1 <= Li <= Ri <= 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|1using 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);}}}}