Codeforces round #257 (Div. 2) Fast Power of B Matrix

B. jzzhu and sequencestime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output

Jzzhu has sorted Ted a kind of sequences, they meet the following property:

You are givenXAndY, Please calculateFNModulo 1000000007 (109? +? 7 ).

Input

The first line contains two integersXAndY(|X| ,? |Y|? ≤? (109). The second line contains a single integerN(1? ≤?N? ≤? 2 · 109 ).

Output

Output a single integer representingFNModulo 1000000007 (109? +? 7 ).

Sample test (s) Input

2 33

Output

1

Input

0 -12

Output

1000000006

Note

In the first sample,F2? =?F1? +?F3, 3? =? 2? +?F3,F3? =? 1.

In the second sample,F2? = ?? -? 1 ;? -? 1 modulo (109? +? 7) equals (109? +? 6 ).

Question

A simple question of the progressive method, I used the Matrix to quickly power it.

The constructor is also simple.

F1 + F3 = F2

F3 = F2-F1

F3 = 1 * F2 +-1 * F1

F2 = 1 * F2 + 0 * F1

So the matrix is

1,-1

1, 0

Then the modulo operation and negative number processing are completed. Negative modulo returns a negative number. Because I am writing a fast power, the number returned at last must be between (-mod, MoD), so I just need to add a mod directly and then take the modulo again.

Sample Code

/*************************************** **************************************** Copyright Notice * copyright (c) 2014 All Rights Reserved * ---- stay hungry Stay Foolish ---- ** @ Author: Shen * @ name: B * @ file: G: \ My source code \ [ACM] competition \ 0719-CF \ B. CPP * @ Date: 2014/07/19 20:57 * @ algorithm: matrix fast power method ************************************ **************************************** **/# include <CST Dio >#include <string >#include <cstring >#include <iostream >#include <algorithm> using namespace STD; typedef long int64; const int maxn = 2; const int maxm = 2; const int mod = 1000000007; struct matrax {int n, m; int64 mat [maxn] [maxm]; matrax (): n (-1 ), M (-1) {} matrax (INT _ n, int _ m): n (_ n), m (_ m) {memset (MAT, 0, sizeof (MAT);} void unit (INT _ s) {n = _ s; M = _ s; For (INT I = 0; I <n; I ++) {for (in T j = 0; j <n; j ++) {mat [I] [J] = (I = J )? 1: 0 ;}}void print () {printf ("n = % d, M = % d \ n", n, m); For (INT I = 0; I <n; I ++) {for (Int J = 0; j <m; j ++) printf ("% 8d", mat [I] [J]); printf ("\ n") ;}}; matrax add_mod (const matrax & A, const matrax & B, const int64 mod) {matrax ans (. n,. m); For (INT I = 0; I <. n; I ++) {for (Int J = 0; j <. m; j ++) {ans. mat [I] [J] = (. mat [I] [J] + B. mat [I] [J]) % mod ;}return ans ;}matrax MUL (const matrax & A, const matrax & B) {matrax ans (. n, B. m); For (INT I = 0; I <. n; I ++) {for (Int J = 0; j <B. m; j ++) {int64 TMP = 0; For (int K = 0; k <. m; k ++) {TMP + =. mat [I] [k] * B. mat [k] [J];} ans. mat [I] [J] = TMP;} return ans;} matrax mul_mod (const matrax & A, const matrax & B, const int mod) {matrax ans (. n, B. m); For (INT I = 0; I <. n; I ++) {for (Int J = 0; j <B. m; j ++) {int64 TMP = 0; For (int K = 0; k <. m; k ++) {TMP + = (. mat [I] [k] * B. mat [k] [J]) % MOD;} ans. mat [I] [J] = TMP % MOD;} return ans;} matrax pow_mod (const matrax & A, int64 K, const int mod) {matrax P (. n,. m), ANS (. n,. m); P = A; ans. unit (. n); If (k = 0) return ans; else if (k = 1) return a; else {While (k) {If (K & 1) {ans = mul_mod (ANS, P, MoD); k --;} else {k/= 2; P = mul_mod (p, p, MoD );}} return ans ;}} int64 X, Y, N, Res; void solve () {CIN >>> x >>> y >> N; If (n = 1) res = x; else if (n = 2) RES = y; else {matrax ans (2, 1); // TMP = CEF ^ (n-2 ); // ans = TMP * beg; // res = ans. mat [0] [0]; matrax CEF (2, 2); CEF. mat [0] [0] = 1; CEF. mat [0] [1] =-1; CEF. mat [1] [0] = 1; CEF. mat [1] [1] = 0; // CEF. print (); matrax beg (2, 1); beg. mat [0] [0] = y; beg. mat [1] [0] = x; matrax TMP (2, 2); TMP = pow_mod (CEF, N-2, MoD); // TMP. print (); ans = mul_mod (TMP, beg, MoD); // ans. print (); Res = ans. mat [0] [0];} If (RES <0) RES + = MOD; cout <res <Endl;} int main () {solve (); return 0 ;}

Codeforces round #257 (Div. 2) Fast Power of B Matrix