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

0
1
2

Sample Output

0
1
42837

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 long

using 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;
}