hdu 4291 A Short problem(矩阵快速幂)

A Short problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2349    Accepted Submission(s): 821

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

012

 

Sample Output

0142837

 

Source

2012 ACM/ICPC Asia Regional Chengdu Online

 

Recommend

liuyiding

 

题解：每层都有一个循环周期，即循环节。

#include<cstdio>#include<algorithm>#include<cstring>#include<iostream>#include<string>#include<vector>#define ll long longusing namespace std;ll n;typedef vector<ll>vec;typedef vector<vec>mat;mat mul(mat &A,mat &B,ll Mod) {    mat C(A.size(),vec(B[0].size()));    for(int i=0; i<A.size(); i++) {        for(int k=0; k<B.size(); k++) {            for(int j=0; j<B[0].size(); j++) {                C[i][j]=(C[i][j]+A[i][k]*B[k][j])%Mod;            }        }    }    return C;}mat pow_mod(mat A,ll x,ll Mod) {    mat B(A.size(),vec(A.size()));    for(int i=0; i<A.size(); i++) {        B[i][i]=1;    }    while(x>0) {        if(x&1)B=mul(B,A,Mod);        A=mul(A,A,Mod);        x>>=1;    }    return B;}int main() {    //freopen("test.in","r",stdin);    ll Mod[3]= {183120,222222224,1000000007};    while(~scanf("%I64d",&n)) {        if(n==0||n==1) {            printf("%I64d\n",n);            continue;        }        mat A(2,vec(2));        ll ans=0;        for(int i=0; i<3; i++) {            A[0][0]=3;            A[0][1]=1;            A[1][0]=1;            A[1][1]=0;            if(n-1<0)continue;            A=pow_mod(A,n-1,Mod[i]);            n=A[0][0]%Mod[i];        }        printf("%I64d\n",n);    }    return 0;}