codeforces 446C DZY Loves Fibonacci Numbers（数学 or 数论+线段树）（两种方法）

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:

1. 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.
2. 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.

 

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 }

View Code

 

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 }

View Code