HDU 4291-A Short problem-循环节+矩阵快速幂

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http://acm.hdu.edu.cn/showproblem.php?pid=4291

　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

g(n)我们可以矩阵快速幂很快求得，但是关键是g(g(n))不好求，g(n)可能很大，不能直接mod，也即g(x)!=g(x%mod)，mod=1e9+7。

打表可以发现在mod=1e9+7下，g(x)=g(x%222222224), 然后把问题转为了求g(g(g(n))%222222224),同理再次求循环节，把问题转为求：

g(g(g(n)%183120)%222222224) ,这样就能分别三次快速幂求到答案

#include<cstdio>#include<algorithm>using namespace std;const int N = 2;  long long  mod;struct Matrix{    long long mat[2][2];} ;Matrix unit_matrix ;long long n, k;Matrix mul(Matrix a, Matrix b) //矩阵相乘{    Matrix res;    for(int i = 0; i < k; i++)        for(int j = 0; j < k; j++)        {            res.mat[i][j] = 0;            for(int t = 0; t < k; t++)            {                res.mat[i][j] += a.mat[i][t] * b.mat[t][j];                res.mat[i][j] %= mod;            }        }    return res;}Matrix pow_matrix(Matrix a, long long m)  //矩阵快速幂{    Matrix res = unit_matrix;    while(m != 0)    {        if(m & 1)            res = mul(res, a);        a = mul(a, a);        m >>= 1;    }    return res;}Matrix get(long long n){    n--;    Matrix ori;    ori.mat[0][0]=1;    ori.mat[0][1]=0;    ori.mat[1][0]=0;    ori.mat[1][1]=0;    Matrix c;    c.mat[0][0]=3;    c.mat[0][1]=1;    c.mat[1][0]=1;    c.mat[1][1]=0;    Matrix ans = pow_matrix(c, n);    ans = mul(ori,ans);    return ans;}int main(){    k=2;    int  i, j, t;    //初始化单位矩阵            //类似快速幂的 ans=1； 如今是ans=单位矩阵    for(i = 0; i < 2; i++)        for(j = 0; j < 2; j++)            unit_matrix.mat[i][j] = 0;    for(i = 0; i < 2; i++)        unit_matrix.mat[i][i] = 1;    while(scanf("%lld",&n)!=EOF)    {        if (n%183120==0)        {            printf("0\n");            continue;        }        mod=183120;        Matrix ans=get(n);        mod=222222224;         if (ans.mat[0][0]==0)        {            printf("0\n");            continue;        }        ans=get(ans.mat[0][0]);        mod=(1e9)+7;        if (ans.mat[0][0]==0)        {            printf("0\n");            continue;        }        ans=get(ans.mat[0][0]);        printf("%lld\n", ans.mat[0][0]);    }    return 0;}

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