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 mqueries, 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).
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.
题目大意：维护一个序列，每次给一段序列加上一个斐波那契数列，或者询问一段序列的和。
思路1：两个斐波那契的定理，用数学归纳法很容易证明：
①定义F[1] = a, F[2] = b, F[n] = F[n - 1] + F[n - 2]（n≥3）。有F[n] = b * fib[n - 1] + a * fib[n - 2]（n≥3），其中fib[i]为斐波那契数列的第 i 项。
②定义F[1] = a, F[2] = b, F[n] = F[n - 1] + F[n - 2]（n≥3）。有F[1] + F[2] + …… + F[n] = F[n + 2] - b。
这题还有一个事实，就是两个上述定义的数列，相加，仍然符合F[n] = F[n - 1] + F[n - 2]的递推公式。
利用这两个定理，用线段树维护序列，线段树的每个结点记录这一段的前两项是什么，预处理好斐波那契数列，便能O(1)地计算出每一个结点中间的数是多少、每一个结点的和。
代码（1513MS）：
1 #include <iostream>
2 #include <algorithm>
3 #include <cstring>
4 #include <cstdio>
5 #include <cmath>
6 using namespace std;
7 typedef long long LL;
8 #define ll (x << 1)
9 #define rr ((x << 1) | 1)
10 #define mid ((l + r) >> 1)
11
12 const int MOD = 1e9 + 9;
13
14 const int MAXN = 300010;
15 const int MAXT = MAXN << 2;
16
17 int f1[MAXT], f2[MAXT], sum[MAXT];
18 int a[MAXN], fib[MAXN];
19 int n, m;
20
21 void init() {
22 fib[1] = fib[2] = 1;
23 for(int i = 3; i <= n + 2; ++i) {
24 fib[i] = fib[i - 1] + fib[i - 2];
25 if(fib[i] >= MOD) fib[i] -= MOD;
26 }
27 }
28
29 int get_fib(int a, int b, int n) {
30 if(n == 1) return a;
31 if(n == 2) return b;
32 return (LL(b) * fib[n - 1] + LL(a) * fib[n - 2]) % MOD;
33 }
34
35 int get_sum(int a, int b, int n) {
36 return (get_fib(a, b, n + 2) - b + MOD) % MOD;
37 }
38
39 void add_fib(int x, int l, int r, int a, int b) {
40 (f1[x] += a) %= MOD;
41 (f2[x] += b) %= MOD;
42 (sum[x] += get_sum(a, b, r - l + 1)) %= MOD;
43 }
44
45 void pushdown(int x, int l, int r) {
46 add_fib(ll, l, mid, f1[x], f2[x]);
47 add_fib(rr, mid + 1, r, get_fib(f1[x], f2[x], mid + 1 - l + 1), get_fib(f1[x], f2[x], mid + 2 - l + 1));
48 f1[x] = f2[x] = 0;
49 }
50
51 void maintain(int x) {
52 sum[x] = (sum[ll] + sum[rr]) % MOD;
53 }
54
55 void build(int x, int l, int r) {
56 if(l == r) {
57 sum[x] = a[l];
58 } else {
59 build(ll, l, mid);
60 build(rr, mid + 1, r);
61 maintain(x);
62 }
63 }
64
65 void update(int x, int l, int r, int a, int b) {
66 if(a <= l && r <= b) {
67 add_fib(x, l, r, fib[l - a + 1], fib[l + 1 - a + 1]);
68 } else {
69 pushdown(x, l, r);
70 if(a <= mid) update(ll, l, mid, a, b);
71 if(mid < b) update(rr, mid + 1, r, a, b);
72 maintain(x);
73 }
74 }
75
76 int query(int x, int l, int r, int a, int b) {
77 if(a <= l && r <= b) {
78 return sum[x];
79 } else {
80 int ret = 0;
81 pushdown(x, l, r);
82 if(a <= mid) (ret += query(ll, l, mid, a, b)) %= MOD;
83 if(mid < b) (ret += query(rr, mid + 1, r, a, b)) %= MOD;
84 return ret;
85 }
86 }
87
88 int main() {
89 scanf("%d%d", &n, &m);
90 for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);
91 init();
92 build(1, 1, n);
93 int op, l, r;
94 while(m--) {
95 scanf("%d%d%d", &op, &l, &r);
96 if(op == 1) update(1, 1, n, l, r);
97 if(op == 2) printf("%d\n", query(1, 1, n, l, r));
98 }
99 }
思路2：按官方题解的说法，有通项公式fib[n] = sqrt(5) / 5 * (((1 + sqrt(5)) / 2) ^ n - ((1 - sqrt(5)) / 2) ^ n)。
5是1e9+9的二次剩余，383008016^2=5(mod 1e9+9)。
利用逆元，可计算出：sqrt(5) / 5、(1 + sqrt(5)) / 2、(1 - sqrt(5)) / 2在模1e9+9意义下的值。
然后，变成用线段树维护两个等比数列。预处理出(1 + sqrt(5)) / 2和(1 - sqrt(5)) / 2的1~n的次方的值，设他们为q，还要求出1-q的逆元（用于计算等比数列的和）。
线段树每个结点记录两个等比数列的首项，跟上面的方法差不多，也是这样维护一个线段树即可。
代码（1996MS）：
1 #include <iostream>
2 #include <algorithm>
3 #include <cstring>
4 #include <cstdio>
5 #include <cmath>
6 using namespace std;
7 typedef long long LL;
8 #define ll (x << 1)
9 #define rr ((x << 1) | 1)
10 #define mid ((l + r) >> 1)
11
12 const int MOD = 1e9 + 9;
13 const int SQRT5 = 383008016;
14
15 const int MAXN = 300010;
16 const int MAXT = MAXN << 2;
17
18 int powa[MAXN], powb[MAXN];
19 int coe, ta, tb, invta, invtb;
20 int fa[MAXT], fb[MAXT], sum[MAXT];
21 int a[MAXN];
22 int n, m;
23
24 int inv(int x) {
25 if(x == 1) return 1;
26 return (LL(MOD - MOD / x) * inv(MOD % x)) % MOD;
27 }
28
29 void init() {
30 coe = inv(SQRT5);
31 ta = (LL(1 + SQRT5) * inv(2)) % MOD;
32 tb = (LL(1 - SQRT5 + MOD) * inv(2)) % MOD;
33 invta = inv(1 - ta + MOD);
34 invtb = inv(1 - tb + MOD);
35 //cout<<coe<<endl<<ta<<endl<<tb<<endl;
36 powa[0] = powb[0] = 1;
37 for(int i = 1; i <= n; ++i) {
38 powa[i] = LL(powa[i - 1]) * ta % MOD;
39 powb[i] = LL(powb[i - 1]) * tb % MOD;
40 }
41 }
42
43 void maintain(int x) {
44 sum[x] = (sum[ll] + sum[rr]) % MOD;
45 }
46
47 void add_fib(int x, int l, int r, int a, int b) {
48 (fa[x] += a) %= MOD;
49 (fb[x] += b) %= MOD;
50 (sum[x] += LL(a) * (1 - powa[r - l + 1] + MOD) % MOD * invta % MOD) %= MOD;
51 (sum[x] -= LL(b) * (1 - powb[r - l + 1] + MOD) % MOD * invtb % MOD) %= MOD;
52 if(sum[x] < 0) sum[x] += MOD;
53 }
54
55 void pushdown(int x, int l, int r) {
56 add_fib(ll, l, mid, fa[x], fb[x]);
57 add_fib(rr, mid + 1, r, LL(fa[x]) * powa[mid + 1 - l] % MOD, LL(fb[x]) * powb[mid + 1 - l] % MOD);
58 fa[x] = fb[x] = 0;
59 }
60
61 void build(int x, int l, int r) {
62 if(l == r) {
63 sum[x] = a[l];
64 } else {
65 build(ll, l, mid);
66 build(rr, mid + 1, r);
67 maintain(x);
68 }
69 }
70
71 void update(int x, int l, int r, int a, int b) {
72 if(a <= l && r <= b) {
73 add_fib(x, l, r, LL(coe) * powa[l - a + 1] % MOD, LL(coe) * powb[l - a + 1] % MOD);
74 } else {
75 pushdown(x, l, r);
76 if(a <= mid) update(ll, l, mid, a, b);
77 if(mid < b) update(rr, mid + 1, r, a, b);
78 maintain(x);
79 }
80 }
81
82 int query(int x, int l, int r, int a, int b) {
83 if(a <= l && r <= b) {
84 return sum[x];
85 } else {
86 int ret = 0;
87 pushdown(x, l, r);
88 if(a <= mid) (ret += query(ll, l, mid, a, b)) %= MOD;
89 if(mid < b) (ret += query(rr, mid + 1, r, a, b)) %= MOD;
90 return ret;
91 }
92 }
93
94 int main() {
95 scanf("%d%d", &n, &m);
96 for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);
97 init();
98 build(1, 1, n);
99 int op, l, r;
100 while(m--) {
101 scanf("%d%d%d", &op, &l, &r);
102 if(op == 1) update(1, 1, n, l, r);
103 if(op == 2) printf("%d\n", query(1, 1, n, l, r));
104 }
105 }
codeforces 446C DZY Loves Fibonacci Numbers（数学 or 数论+线段树）,布布扣,bubuko.com