Tenth multi-school HDU 3936 fib query (the nature of the Fibonacci series and the ologn matrix acceleration Multiplication Algorithm)

The general formula of the Fibonacci series is F (n) = (1/√ 5) * {[(1 + √ 5)/2] ^ (n + 1) -[(1-√ 5)/2] ^ (n + 1 )}

Nature:

 
1. F (0) + F (1) + F (2) +... + F (n) = f (n + 2)-1. 2. F (1) + f (3) + f (5) +... + F (2n-1) = f (2n ). 3. F (2) + f (4) + f (6) +... + F (2n) = f (2n + 1)-1. 4. [F (0)] ^ 2 + [F (1)] ^ 2 +... + [F (n)] ^ 2 = f (n) · F (n + 1 ). 5. F (0)-f (1) + F (2 )-... + (-1) ^ N · F (n) = (-1) ^ N · [F (n + 1)-f (n)] + 1. 6. F (m + N-1) = f (S-1) · F (n-1) + f (m) · F (n ). This can be usedProgramCompile a program whose time complexity is only O (log n. 7. [F (n)] ^ 2 = (-1) ^ (n-1) + f (n-1) · F (n + 1 ). 8. F (2n-1) = [F (n)] ^ 2-[F (n-2)] ^ 2. 9.3f (n) = f (n + 2) + f (n-2 ). 10. F (2n-2m-2) [F (2n) + f (2n + 2)] = f (2 m + 2) + f (4n-2m) [n> M ≥-1, n ≥ 1] 11.f( 2n + 1) = [F (n)] ^ 2 + [F (n + 1)] ^ 2.
 

 
Integer Division and prime generation of the Fibonacci seriesEach 3 numbers has only one divisible by 2. Each 4 numbers has only one divisible by 3. Each 5 numbers has only one divisible by 5, each 6 numbers has only one integer divided by 8, and each 7 numbers has only one integer divided by 13. Each 8 numbers has only one integer divided by 21, each 9 numbers have only one divisible by 34 ,....... we can see that 5th, 7, 11, 13, 17, and 23 are prime numbers: 5, 13, 89,233,159, 19th, 28657 (bits are not)
 

The number of digits in the Fibonacci sequence: a 60-step cycleListen 35, 83145, 94370,77415, 61785.38190, 99875,27965, 16730,31045, 49325,72910...
 

Question:

This question can be exploited in Nature 4 and 6

Define sum [I] = sigma (P [J]) 1 <= j <= I.

P [I] = f [2 * I-1] ^ 2 + F [2 * I] ^ 2.

According to formula 4, sum [I] = f [2 * I] * f [2 * I + 1].

Then sum [R]-sum L-1] can be obtained directly.

This is the first time that I used matrix concatenation to pre-process f [2 ^ N].

HDU ran 78 Ms 11th. It is estimated that the first 10 running 62 were all pasted with the same template.

# Include <cstdio> # include <cstring> typedef long ll; const ll mod = 1000000007; ll mtrx [60] [2] [2]; void debug (ll a [] [2]) {printf ("% LLD \ n", a [0] [0], A [0] [1], a [1] [0], a [1] [1]);} void pre_pro () {memset (mtrx, 0, sizeof (mtrx); mtrx [0] [0] [0] = 1; mtrx [0] [0] [1] = 1; mtrx [0] [1] [0] = 1; mtrx [0] [1] [1] = 0; For (int t = 0; t <60; ++ T) for (INT I = 0; I <2; ++ I) for (Int J = 0; j <2; ++ J) {for (int K = 0; k <2; ++ K) mtrx [t + 1] [I] [J] + = mtrx [T] [I] [k] * mtrx [T] [k] [J]; mtrx [t + 1] [I] [J] % = mod ;}} ll fib (LL A) {A --; ll mat [2] [2] = {1, 0, 0, 1}; ll TMP [2] [2]; for (INT p = 0; A >>= 1, ++ p) {If (! (A & 1) continue; TMP [0] [0] = mat [0] [0]; TMP [0] [1] = mat [0] [1]; TMP [1] [0] = mat [1] [0]; TMP [1] [1] = mat [1] [1]; memset (MAT, 0, sizeof (MAT); For (INT I = 0; I <2; ++ I) for (Int J = 0; j <2; ++ J) {for (int K = 0; k <2; ++ K) mat [I] [J] + = mtrx [p] [I] [k] * TMP [k] [J]; mat [I] [J] % = mod ;}// debug (MAT); Return mat [0] [0];} ll sum (LL) {if (a = 0) return 0; else return fib (2 * A) * fib (2 * A + 1) % MOD;} int main () {pre_pro (); int CAS; // freopen ("in. in "," r ", stdin); // freopen (" out.txt "," W ", stdout); // For (INT I = 0; I <20; ++ I) debug (mtrx [I]); scanf ("% d", & CAS); While (CAS --) {ll l, R; scanf ("% i64d % i64d", & L, & R); printf ("% d \ n", (sum (R)-sum L-1) + mod) % mod);} return 0 ;}