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Portal

Portal1: Codeforces

Portal2: Luogu

Description

In mathematical terms, the sequence \(F_n\) of Fibonacci numbers is defined by the recurrence relation

\[F_1 = 1; F_2 = 1; F_n = F_n - 1 + F_n - 2 (n > 2)\]

DZY loves Fibonacci numbers very much. Today DZY gives you an array consisting of \(n\) integers: \(a1, a2, \cdots , an\). Moreover, there are \(m\) queries, each query has one of the two types:

1. Format of the query "1 l r". In reply to the query, you need to add \(F_i - l + 1\) to each element ai, where \(l \le i \le r\).

2. Format of the query "2 l r". In reply to the query you should output the value of modulo \(1000000009 (10^9 + 9)\).

Help DZY reply to all the queries.

Input

The first line of the input contains two integers \(n\) and \(m (1 \le n, m \le 300000)\). The second line contains \(n\) integers \(a_1, a_2, \cdots , a_n (1 \le ai \le 10^9)\) — initial array \(a\) .

Then, \(m\) lines follow. A single line describes a single query in the format given in the statement. It is guaranteed that for each query inequality \(1 \le l \le r \le n\) holds.

Output

For each query of the second type, print the value of the sum on a single line.

Sample Input

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

Sample Output

17
12

Sample Explain

After the first query, \(a = [2, 3, 5, 7]\).

For the second query, \(sum = 2 + 3 + 5 + 7 = 17\).

After the third query, \(a = [2, 4, 6, 9]\).

For the fourth query, \(sum = 2 + 4 + 6 = 12\).

Description in Chinese

题目让我们求给你一个序列，支持区间加Fibonacci数列前r - l + 1项和查询区间和。

Solution

一些约定：把斐波那契数列的前两个数\(F_1 = 1, F_2 = 1\)换成另两个数，仍满足\(F_n = F_{n - 1} + F_{n - 2}(n > 2)\)的数列称为广义斐波那契数列。

Fibonacci数列有一些性质：

性质\(1\). \(F_n = (\sum^{n - 2}_{i = 1}{F_i}) + F_2(n > 2)\)；

证明如下：

首先将前几项Fibonacci数列展开。

F(1) = 1
F(2) = 1
F(3) = F(1) + F(2)
F(4) = F(2) + F(3) = F(2) + F(1) + F(2)
F(5) = F(3) + F(4) = F(3) + F(2) + F(1) + F(2)
F(6) = F(4) + F(5) = F(4) + F(3) + F(2) + F(1) + F(2)
......

在\(F_n = F_{n - 1} + F_{n - 2}\)中，我们可以把\(F_{n - 1}\)按式子展开，可得\(F_n = \sum^{n - 3}_{i = 1} + F_2 + F_{n - 2}\)，即\(F_n = (\sum^{n - 2}_{i = 1}{F_i}) + F_2(n > 2)\)，跟原式一模一样，故原式正确性得证。

性质\(2\). 一个广义斐波那契数列数列\(f_i\)， 当\(f_1 = x, f_2 = y\)时，则有\(f_n = x \times f_{n - 1} + y \times f_{n - 2}\)

证明如下：

这个性质与性质1类似，证明方法也与性质1类似，列举几个：

f(1) = x
f(2) = y
f(3) = f(1) + f(2) = x × F(1)
f(4) = f(2) + f(3) = x × F(1) + y × F(2)
f(5) = f(3) + f(4) = x × F(2) + y × F(3)
f(6) = f(4) + f(5) = x × F(3) + y × F(4)
......

把上述规律推广到代数式：

\[f_n = f_{n - 1} + f_{n - 2} \\\\ \quad = x \times f_{n - 2} + y \times f_{n - 3} + x \times f_{n - 3} + y \times f_{n - 4} \\\\ \quad = x \times (f_{n - 2} + f_{n - 3}) + y \times (f_{n - 3} + f_{n - 4}) \\\\ \quad = x \times f_{n - 1} + y \times f_{n - 2}\]

证毕。

性质\(3\): 任意两段不同的广义斐波那契数列段相加（逐项相加），所得的数列任然是广义斐波那契数列。

这个性质易证。

这题我们维护一棵线段树，线段树需要维护\(L\)至\(R\)区间的广义斐波那契数列的第一项，第二项与区间的和。

下传标记时，我们可以在左区间加广义斐波那契数列的前两项，在右区间可以求出总和再加上总和就行了，时间复杂\(\text{O(n log n)}\)。

Code

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>using namespace std;typedef long long LL;
const int MAXN = 300005, MAXM = 1200005, mod = 1e9 + 9;
struct node {int c1, c2, sum;
} tree[MAXM];
int n, m, opt, x, y, a[MAXN], f[MAXN];
inline int add(int x, int y) {//两项相加并取模int ret = x + y;if (ret < 0) return ret += mod; else return ret % mod;
}
inline int calc1(int x, int y, int len) {//计算斐波那契if (len == 1) return x; elseif (len == 2) return y; else return ((LL)x * f[len - 2] + (LL)y * f[len - 1]) % mod;
}
inline int calc2(int x, int y, int len) {//计算总和if (len == 1) return x; elseif (len == 2) return add(x, y); else return add(calc1(x, y, len + 2), -y);
}
inline void pushup(int rt) {tree[rt].sum = add(tree[rt << 1].sum, tree[rt << 1 | 1].sum);
}
inline void pushdown(int rt, int l, int r) {//下传标记if (tree[rt].c1) {int mid = l + r >> 1;tree[rt << 1].c1 = add(tree[rt << 1].c1, tree[rt].c1);tree[rt << 1].c2 = add(tree[rt << 1].c2, tree[rt].c2);tree[rt << 1].sum = add(tree[rt << 1].sum, calc2(tree[rt].c1, tree[rt].c2, mid - l + 1));int x = calc1(tree[rt].c1, tree[rt].c2, mid - l + 2), y = calc1(tree[rt].c1, tree[rt].c2, mid - l + 3);tree[rt << 1 | 1].c1 = add(tree[rt << 1 | 1].c1, x);tree[rt << 1 | 1].c2 = add(tree[rt << 1 | 1].c2, y);tree[rt << 1 | 1].sum = add(tree[rt << 1 | 1].sum, calc2(x, y, r - mid));tree[rt].c1 = 0; tree[rt].c2 = 0;}
}
inline void update(int rt, int l, int r, int ansl, int ansr) {//线段树区间更新if (ansl <= l && r <= ansr) {tree[rt].c1 = add(tree[rt].c1, f[l - ansl + 1]);tree[rt].c2 = add(tree[rt].c2, f[l - ansl + 2]);tree[rt].sum = add(tree[rt].sum, calc2(f[l - ansl + 1], f[l - ansl + 2], r - l + 1));return ;}pushdown(rt, l, r);int mid = l + r >> 1;if (ansl <= mid) update(rt << 1, l, mid, ansl, ansr);if (ansr > mid) update(rt << 1 | 1, mid + 1, r, ansl, ansr);pushup(rt);
}
inline int query(int rt, int l, int r, int ansl, int ansr) {//线段树区间查询int ret = 0;if (ansl <= l && r <= ansr) {ret = tree[rt].sum;return ret;}pushdown(rt, l, r);int mid = l + r >> 1;if (ansl <= mid) ret = add(ret, query(rt << 1, l, mid, ansl, ansr));if (ansr > mid) ret = add(ret, query(rt << 1 | 1, mid + 1, r, ansl, ansr));return ret;
}
int main() {scanf("%d%d", &n, &m);for (int i = 1; i <= n; i++) {scanf("%d", &x);a[i] = add(a[i - 1], x);}f[1] = 1; f[2] = 1;for (int i = 3; i <= n + 2; i++)f[i] = add(f[i - 1], f[i - 2]);for (int i = 1; i <= m; i++) {scanf("%d%d%d", &opt, &x, &y);if (opt == 1) update(1, 1, n, x, y); else printf("%d\n", add(query(1, 1, n, x, y), a[y] - a[x - 1]));}return 0;
}