In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation
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).
Help DZY reply to all the queries.
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.
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 ", 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 = frac{ sqrt 5 } 5 * (( frac {1 + sqrt 5} 2) ^ n - ( frac {1 - sqrt 5} 2) ^ n) $。
5是1e9+9的二次剩余，383008016^2=5(mod 1e9+9)。
利用逆元，可计算出：$ frac {sqrt 5} 5、frac {1 + sqrt 5} 2、 frac {1 - sqrt 5} 2 $在模1e9+9意义下的值。
然后，变成用线段树维护两个等比数列。预处理出$frac {1 + sqrt 5} 2$和$frac {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 ", query(1, 1, n, l, r)); 104 } 105 }