【HDU】4291 A Short problem 矩阵快速幂
2014-07-04 21:05
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A Short problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1486 Accepted Submission(s): 540
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 should solve for
g(g(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 Input
0 1 2
Sample Output
0 1 42837
Source
2012 ACM/ICPC Asia Regional
Chengdu Online
传送门:【HDU】4291 A Short problem
题目分析:
mod1 = 1e9 + 7 ,g(x) = 3 * g(x - 1) + g(x - 2)
首先暴力找循环节,先对外层g(x)%mod1,设a = g(1),b = g(0),则如果循环到i的时候g(i) == a && g(i-1) == b则x的循环节就是mod2 = i - 1。
那么对任意的x%mod2后带入g函数中与原样带入是等价的。
接下来设x = f(y),然后对该层f(y)%mod2,同样设a = f(1),b = f(0),则同样循环到j的时候满足f(j) == a && f(j-1) == b则y的循环节是mod3 = j - 1。
那么对于任意的y%mod3后带入f函数中与原样带入依旧等价。
所以我们得到了两个循环节再加上最初的mod1就构成了三重循环节。
所以,g(g(g(n)))%mod1 = g(g(g(n)%mod3)%mod2)%mod1。
其中
mod1 = 1000000007 ;
mod2 = 222222224 ;
mod3 = 183120 ;
问题得到完美解决~
代码如下:
#include <stdio.h> #include <string.h> #include <algorithm> using namespace std ; #define REP( i , n ) for ( int i = 0 ; i < n ; ++ i ) typedef long long LL ; const int MOD1 = 1000000007 ; const int MOD2 = 222222224 ; const int MOD3 = 183120 ; const int MAXN = 2 ; struct Matrix { int mat[MAXN][MAXN] ; int N ; Matrix () {} Matrix ( int n ) { N = n ; REP ( i , N ) REP ( j , N ) mat[i][j] = 0 ; } void init () { REP ( i , N ) mat[i][i] = 1 ; } void build () { mat[0][0] = 3 ; mat[0][1] = 1 ; mat[1][0] = 1 ; mat[1][1] = 0 ; } } ; Matrix E , A ; Matrix mul ( Matrix a , Matrix b , int mod ) { Matrix res = Matrix ( a.N ) ; REP ( i , a.N ) REP ( j , a.N ) REP ( k , a.N ) res.mat[i][j] = ( res.mat[i][j] + ( LL ) a.mat[i][k] * b.mat[k][j] % mod ) % mod ; return res ; } LL pow ( LL k , int mod ) { Matrix res = E , tmp = A ; while ( k ) { if ( k & 1 ) res = mul ( res , tmp , mod ) ; tmp = mul ( tmp , tmp , mod ) ; k >>= 1 ; } return res.mat[0][0] ; } void work () { LL k ; E = Matrix ( 2 ) ; A = Matrix ( 2 ) ; E.init () ; A.build () ; while ( ~scanf ( "%I64d" , &k ) ) { if ( k >= 2 ) k = pow ( k - 1 , MOD3 ) ; if ( k >= 2 ) k = pow ( k - 1 , MOD2 ) ; if ( k >= 2 ) k = pow ( k - 1 , MOD1 ) ; printf ( "%I64d\n" , k ) ; } } int main () { work () ; return 0 ; }
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