HDU 4291 a short problem 37th ACM/ICPC Chengdu division Network Competition 1004 questions (find the rule, take the model to find the cycle Section)

A Short Problem

Time Limit: 2000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)
Total submission (s): 344 accepted submission (s): 131

Problem description according to a research, VIM users tend to have shorter fingers, compared with Emacs users.
Hence they prefer problems short, too. Here is a short one:
Given N (1 <=n <= 1018), You shoshould solve
G (N) mod 109 + 7
Where
G (n) = 3G (n-1) + g (n-2)
G (1) = 1
G (0) = 0

 

Input there are several test cases. For each test case there is an integer N in a single line.
Please process until EOF (end of file ).

 

Output for each test case, please print a single line with a integer, the corresponding answer to this case.

 

Sample input0 1 2

 

Sample output0 1 42837

 

Source2012 ACM/ICPC Asia Regional Chengdu online

 

Recommendliuyiding I didn't even make such a water question competition... Sorry !!!! In fact, as long as the model is used, you can find the loop section. The first time is MoD = 1000000007 find the loop section is 222222224
The second time is MoD = 222222224, find the cycle section 183120 You can find the cyclic festival brute force, and the matrix multiplication method is slower. Attached cycle searching sectionProgram: View code

# Include <stdio. h> Const   Long   Long MoD = 222222224 ; //  The first time is MoD = 1000000007 find the loop section is 222222224  //  The second time is MoD = 222222224, find the cycle section 183120 Int  Main (){  Long   Long  A, B; = 1  ; B = 3  ;  For ( Int I = 1 ; I ++ ){  If (A = 0 & B =1  ) {Printf (  "  % D \ n  "  , I );  Break  ;}  Long   Long C = 3 * B + A; c % = MOD; = B; B = C ;} Return   0  ;} 

 

The following is a program.

 //  1004 # Include <stdio. h> # Include <Iostream> # Include <Map> # Include < Set > # Include <Algorithm> # Include < String . H> # Include <Stdlib. h> Using   Namespace  STD;  Const   Long   Long MoD = 1000000007  ;  Const   Long   Long Mod2 = 222222224  ;  Const   Long   Long Mod3 =183120  ;  Const   Int Maxn = 10  ;  Struct  Matrix {  Long   Long  Mat [maxn] [maxn];  Int  N, m;}; matrix MUL (matrix A, matrix B,  Long   Long  M) {matrix ret; ret. n =A. N; ret. m = B. m;  For ( Int I = 0 ; I <A. N; I ++ )  For ( Int J = 0 ; J <B. m; j ++ ) {Ret. Mat [I] [J] = 0  ;  For ( Int K =0 ; K <a. m; k ++ ) {Ret. Mat [I] [J] + = (A. Mat [I] [k] * B. Mat [k] [J] % M); ret. Mat [I] [J] % = M ;}}  Return  RET;} matrix POW (matrix,  Long   Long N, Long   Long  M ){  If (N = 1 )Return  A; matrix RET = A; matrix temp = A;  For ( Int I = 0 ; I <A. N; I ++ )  For ( Int J = 0 ; J <A. N; j ++ ){  If (I = J) ret. Mat [I] [J] = 1 ;  Else Ret. Mat [I] [J] = 0  ;}  While  (N ){  If (N & 1 ) Ret = Mul (Ret, temp, m); temp = Mul (temp, temp, m); n >>= 1  ;}  Return  RET ;} Int  Main (){  //  Freopen ("D. In", "r", stdin );  //  Freopen ("D. Out", "W", stdout );  Matrix A; A. N = A. m = 2  ; A. Mat [  0 ] [ 0 ] = 3  ; A. Mat [  0 ] [ 1 ] =1  ; A. Mat [  1 ] [ 0 ] = 1  ; A. Mat [  1 ] [ 1 ] = 0  ; Matrix F0; f0.n = 2  ; F0.m = 1  ; F0.mat [  0 ] [0 ] = 1  ; F0.mat [  1 ] [ 0 ] = 0  ;  Long   Long  N; matrix TMP; matrix temp;  While (Scanf ( "  % I64d \ n  " , & N )! = EOF) {temp =Pow (A, N, mod3); TMP = Mul (temp, F0, mod3 );  Long   Long Tt = TMP. Mat [ 1 ] [ 0  ]; Temp = Pow (A, TT, mod2); TMP = Mul (temp, F0, mod2); TT = TMP. Mat [ 1 ] [ 0  ]; Temp = Pow (A, TT, MoD); TMP =Mul (temp, F0, MoD); TT = TMP. Mat [ 1 ] [ 0  ]; Printf (  "  % I64d \ n  "  , TT );}  Return   0  ;}