hdu 4291 A Short problem

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本文讨论了如何高效解决复杂递归序列求模问题，通过矩阵快速幂技巧优化计算过程，适用于需要频繁求解特定序列模运算的场景。

A Short problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 852 Accepted Submission(s): 335

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 <= 10 ¹⁸), You should solve for
g(g(g(n))) mod 10 ⁹ + 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

Sample Output

Source

2012 ACM/ICPC Asia Regional Chengdu Online

代码：

#include<cstdio>
#define ll __int64
int mod;
struct Matrix
{
    ll m[2][2];
    void init()
    {
        m[0][0]=3;m[0][1]=1;
        m[1][0]=1;m[1][1]=0;
    }
}a;
Matrix mul(Matrix a,Matrix b)
{
    Matrix c;
    int i,j,k;
    for(i=0;i<2;i++)for(j=0;j<2;j++)
    {
        c.m[i][j]=0;
        for(k=0;k<2;k++)
            c.m[i][j]=(a.m[i][k]*b.m[k][j]%mod+c.m[i][j])%mod;
    }
    return c;
}
Matrix mpow(Matrix a,ll n)
{
    Matrix c;
    int i,j;
    for(i=0;i<2;i++)for(j=0;j<2;j++)c.m[i][j]=(i==j);
    for(;n;n>>=1)
    {
        if(n&1)c=mul(c,a);
        a=mul(a,a);
    }
    return c;
}
ll solve(ll n,ll m)
{
    a.init();
    mod=m;
    a=mpow(a,n);
    return a.m[0][1];
}
int main()
{
    ll n;
    while(~scanf("%I64d",&n))
        printf("%I64d\n",solve(solve(solve(n,183120),222222224),1000000007));
    return 0;
}