CF446C [DZY loves Fibonacci]

Description

Transmission Gate

你需要维护一个长度为\(n \leq 300000\) 的数列，兹词两个操作：

1.给一个区间加上一个fibonacci数列，规定\(f[0] = 0, f[1] = 1, f[2] = 1\)

2.查询一段区间的和。对1e9+9取模

操作个数m不超过300000.

Solution

​ 这一题其实要考虑Fibonacci数列的两个性质:

​ (i)\(\sum_{i = 1}^{n} fib(i) = f(n + 2) - 1\)

​ (ii)令\(S_i = S_{i - 1} + S_{i - 2}\), 其中\(S_1 = a, S_2 = b\)

​ 那么\(S_i = aFib(i - 1) + bFib(i - 2)\)

​ 那么就很好做了， 在线段树中， 我们要只要记数列的前两项就可以方便的对数列进行求和，pushdown等操作.

​ 这题细节比较多，相关部分见代码。

Code

#include<bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for(int i = (a), i##_end_ = (b); i <= i##_end_; ++i)
#define drep(i, a, b) for(int i = (a), i##_end_ = (b); i >= i##_end_; --i)
#define clar(a, b) memset((a), (b), sizeof(a))
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define Debug(s) debug("The massage in line %d, Function %s: %s\n", __LINE__, __FUNCTION__, s)
typedef long long LL;
typedef long double LD;
const int BUF_SIZE = (int)1e6 + 10;
struct fastIO {
	char buf[BUF_SIZE], buf1[BUF_SIZE];
	int cur, cur1;
	FILE *in, *out;
	fastIO() {
		cur = BUF_SIZE, in = stdin, out = stdout;
		cur1 = 0;
	}
	inline char getchar() {
		if(cur == BUF_SIZE) fread(buf, BUF_SIZE, 1, in), cur = 0;
		return *(buf + (cur++));
	}
	inline void putchar(char ch) {
		*(buf1 + (cur1++)) = ch;
		if (cur1 == BUF_SIZE) fwrite(buf1, BUF_SIZE, 1, out), cur1 = 0;
	}
	inline int flush() {
		if (cur1 > 0) fwrite(buf1, cur1, 1, out);
		return cur1 = 0;
	}
}IO;
#define getchar IO.getchar
#define putchar IO.putchar
int read() {
	char ch = getchar();
	int x = 0, flag = 1;
	for(;!isdigit(ch); ch = getchar()) if(ch == '-') flag *= -1;
	for(;isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
	return x * flag;
}
void write(int x) {
	if(x < 0) putchar('-'), x = -x;
	if(x >= 10) write(x / 10);
	putchar(x % 10 + 48);
}
void putString(char s[], char EndChar = '\n') {
	rep(i, 0, strlen(s) - 1) putchar(*(s + i));
	if(~EndChar) putchar(EndChar);
}

#define Maxn 300009
const LL Mod = 1000000009;
int n, a[Maxn], m; LL f[Maxn];
struct node { LL x, y; };
LL FibDelta(LL F1, LL F2, LL len) {
	LL res = 0;
	if(len >= 1) (res += F1) %= Mod;
	if(len >= 2) (res += F2) %= Mod;
	if(len >= 3) (res += (F2 * (f[len + 1] - 2ll) % Mod + F1 * (f[len] - 1ll) % Mod) % Mod) %= Mod;
	return res;
}// It needs to analysis in different conditions
node FibForward(node s, int len) {
	LL c = s.x * f[len] % Mod + s.y * f[len + 1] % Mod, d = s.x * f[len + 1] % Mod + s.y * f[len + 2] % Mod;
	return (node){c % Mod, d % Mod};
}
namespace SGMT_tree {
	LL tree[Maxn << 2], beg[Maxn << 2][2], val[Maxn << 2];
#define lc(x) (x) << 1
#define rc(x) (x) << 1 | 1
#define ls rt << 1, l, mid
#define rs rt << 1 | 1, mid + 1, r
	void pushup(int rt) { tree[rt] = (1ll * tree[lc(rt)] + 1ll * tree[rc(rt)]) % Mod; }
	void pushdown(int rt, int l, int r) {
		int mid = (l + r) >> 1;
		if(val[rt]) {
			LL &a = beg[rt][0], &b = beg[rt][1];
			(beg[lc(rt)][0] += a) %= Mod, (beg[lc(rt)][1] += b) %= Mod;
			(tree[lc(rt)] += FibDelta(a, b, mid - l + 1)) %= Mod;
			val[lc(rt)] = 1;

			LL c = a * f[mid - l] + b * f[mid - l + 1], d = a * f[mid - l + 1] + b * f[mid - l + 2];
			c %= Mod, d %= Mod;
			(beg[rc(rt)][0] += c) %= Mod, (beg[rc(rt)][1] += d) %= Mod;
			(tree[rc(rt)] += FibDelta(c, d, r - mid)) %= Mod;
			val[rc(rt)] = 1;
			beg[rt][0] = beg[rt][1] = 0;
			val[rt] = 0;
		}
	}
	void build(int rt, int l, int r) {
		if(l == r) {
			tree[rt] = a[l];
			return ;
		}
		int mid = (l + r) >> 1;
		build(ls), build(rs);
		pushup(rt);
	}
	void modify(int rt, int l, int r, int x, int y, int p1, int p2) {
		if(x <= l && r <= y) {
			(beg[rt][0] += p1) %= Mod, (beg[rt][1] += p2) %= Mod;
			(tree[rt] += FibDelta(p1, p2, r - l + 1)) %= Mod;
			val[rt] = 1;
			return ;
		}

		int mid = (l + r) >> 1;
		pushdown(rt, l, r);

		if(y <= mid)  modify(ls, x, y, p1, p2);
		else if(mid + 1 <= x)  modify(rs, x, y, p1, p2);
		else {
			modify(ls, x, mid, p1, p2); /**/
			node z = FibForward((node){p1, p2}, mid - x);/**/
			//F[i] -> F[i + mid - x]: p1 -> z.x, p2 -> z.y
			modify(rs, mid + 1, y, z.x, z.y);/*The InterVal needs to move*/
		}
		pushup(rt);
	}
	int query(int rt, int l, int r, int x, int y) {
		if(x <= l && r <= y) return tree[rt];

		int mid = (l + r) >> 1; pushdown(rt, l, r);

		if(y <= mid) return query(ls, x, y);
		else if(mid + 1 <= x) return query(rs, x, y);
		else return (1ll * query(ls, x, y) + 1ll * query(rs, x, y)) % Mod;
	}
#undef lc
#undef rc
#undef ls
#undef rs
}
namespace INIT {
	void Main() {
		n = read(), m = read();

		f[1] = 1; f[2] = 1;
		rep(i, 3, n + 4) f[i] = (f[i - 1] * 1ll + f[i - 2]) % Mod;

		rep(i, 1, n) a[i] = read();

		SGMT_tree :: build(1, 1, n);
	}
}
namespace SOLVE {
	void Main() {
		rep(i, 1, m) {
			int opt = read();

			if(opt == 1) {
				int x = read(), y = read();
				SGMT_tree :: modify(1, 1, n, x, y, 1, 1);
			}
			if(opt == 2) {
				int x = read(), y = read();
				write(SGMT_tree :: query(1, 1, n, x, y)), putchar('\n');
			}
		}
	}
}
int main() {
	freopen("CF446C.in", "r", stdin);
	freopen("CF446C.out", "w", stdout);

	INIT :: Main();
	SOLVE :: Main();
#ifdef Qrsikno
	debug("\nRunning time: %.3lf(s)\n", clock() * 1.0 / CLOCKS_PER_SEC);
#endif
	return IO.flush();
}