【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

1. Fn=Fn−2+Fn−1
2. 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

1. Hn=a∗Fn−2+b∗Fn−1
2. 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 longconst int N = 333333;const ll mod = 1000000009ll;int n, m;struct node {    int l, r;    ll sum, f1, f2;}t[N * 6];ll a[N], F[N];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.