【Codeforces Round #FF (Div. 2)】E. DZY Loves Fibonacci Numbers

题目

E. DZY Loves Fibonacci Numbers

Background

In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation

F1 = 1; F2 = 1; Fn = Fn - 1 + Fn - 2 (n > 2).

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

Format of the query “1 l r”. In reply to the query, you need to add Fi - l + 1 to each element ai, where l ≤ i ≤ r.

Format of the query “2 l r”. In reply to the query you should output the value of modulo 1000000009 (109 + 9).

Help DZY reply to all the queries.

Input

The first line of the input contains two integers n and m (1 ≤ n, m ≤ 300000). The second line contains n integers a1, a2, …, an (1 ≤ ai ≤ 109) — 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 ≤ l ≤ r ≤ n holds.

Output

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

Sample test(s) input

4 4

1 2 3 4

1 1 4

2 1 4

1 2 4

2 1 3

output

17

12

Note

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.

思路

这是一个显然的区间修改与查询问题，想到使用线段树

但是每一个区间修改的值不一样，怎么办呢TAT

F1=F2=1的Fibonacci数列有以下性质

Fn=Fn−2+Fn−1

F1+F2+…+Fn=Fn+2−1

而且两个Fibonacci数列逐项相加仍然示意个Fibonacci数列

将上述的两个性质推广到H1=a H2=b的Fibonacci数列，得到以下性质

Hn=a∗Fn−2+b∗Fn−1

H1+H2+…+Hn=Hn+2−b

这样就解决了区间修改问题，对于每一个区间记tag就改为维护一个长度固定的Fibonacci数列，对每一个区间记录这个数列的前两项，当pushdown时也可以O(1)地修改子区间的前两项

代码

#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for(int i = (a); i <= (b); i++)
#define red(i, a, b) for(int i = (a); i >= (b); i--)
#define ll long long

const int N = 333333;
const ll mod = 1000000009ll;
int n, m;
struct node {
int l, r;
ll sum, f1, f2;
}t[N * 6];
ll a
, F
;

inline ll read() {
ll x = 0;
char c = getchar();
while(!isdigit(c)) c = getchar();
while(isdigit(c)) {
x = x * 10ll + c - '0';
c = getchar();
}
return x;
}

void calc_Fibonacci(int n) {
F[1] = 1; F[2] = 1;
rep(i, 3, N) F[i] = (F[i - 1] + F[i - 2]) % mod;
}

ll calc(ll a, ll b, ll n) {
if (n == 1) return a;
else if (n == 2) return b;
else return (a * F[n - 2] % mod + b * F[n - 1] % mod) % mod;
}

ll calc_range(ll a, ll b, ll n) {
if (n == 1) return a;
else if (n == 2) return (a + b) % mod;
else return (calc(a, b, n + 2) - b + mod) % mod;
}

void build(int x, int l, int r) {
int mid = (l + r) / 2, lc = x * 2, rc = x * 2 + 1;
t[x].l = l; t[x].r = r;
if (l == r) {
t[x].sum = a[l];
t[x].f1 = t[x].f2 = 0;
return;
}
build(lc, l, mid);
build(rc, mid + 1, r);
t[x].sum = (t[lc].sum + t[rc].sum) % mod;
t[x].f1 = t[x].f2 = 0;
}

void pushdown(int x) {
int l = t[x].l, r = t[x].r;
int mid = (l + r) / 2, lc = x * 2, rc = x * 2 + 1;
t[lc].f1 = (t[lc].f1 + t[x].f1) % mod;
t[lc].f2 = (t[lc].f2 + t[x].f2) % mod;
t[lc].sum += calc_range(t[x].f1, t[x].f2, mid - l + 1);
t[lc].sum %= mod;
t[rc].f1 = (t[rc].f1 + calc(t[x].f1, t[x].f2, mid - l + 2)) % mod;
t[rc].f2 = (t[rc].f2 + calc(t[x].f1, t[x].f2, mid - l + 3)) % mod;
t[rc].sum += calc_range(t[x].f1, t[x].f2, r - l + 1) - calc_range(t[x].f1, t[x].f2, mid - l + 1);
t[rc].sum = (t[rc].sum + mod) % mod;
t[x].f1 = t[x].f2 = 0;
}

void update(int x, int l, int r, int ql, int qr) {
int mid = (l + r) / 2, lc = x * 2, rc = x * 2 + 1;
if (l >= ql && r <= qr) {
t[x].f1 = (t[x].f1 + F[l - ql + 1]) % mod;
t[x].f2 = (t[x].f2 + F[l - ql + 2]) % mod;
t[x].sum += calc_range(F[l - ql + 1], F[l - ql + 2], r - l + 1);
t[x].sum %= mod;
return;
}
pushdown(x);
if (ql <= mid) update(lc, l, mid, ql, qr);
if (qr > mid) update(rc, mid + 1, r, ql, qr);
t[x].sum = (t[lc].sum + t[rc].sum) % mod;
return;
}

ll query(int x, int l, int r, int ql, int qr) {
int mid = (l + r) / 2, lc = x * 2, rc = x * 2 + 1;
if (l >= ql && r <= qr) return t[x].sum;
pushdown(x);
ll ans_left = 0, ans_right = 0;
if (ql <= mid) ans_left = query(lc, l, mid, ql, qr);
if (qr > mid) ans_right = query(rc, mid + 1, r, ql, qr);
return (ans_left + ans_right) % mod;
}

int main() {
n = read(); m = read();
calc_Fibonacci(n);
rep(i, 1, n) a[i] = read();
build(1, 1, n);
rep(i, 1, m) {
int tag, l, r;
tag = read();
l = read();
r = read();
if (tag == 1) update(1, 1, n, l, r);
else printf("%lld\n", query(1, 1, n, l, r));
}
return 0;
}

尾声

我的运行时间成功跑到了Accepted中的倒数第三

不过代码长度倒是排在第一面的

调试时间之长源自于把%mod写成%n，也是醉

这次的思路和公式写的很好看

终于知道Markdown怎么调字体了

SR真是Hentai呢

End.