【ACMICPC Chengdu Online】1004 数论，类Fib

题意：

公式g[n] = 3 * g[n-1] + g[n-2]

          求： g(g(g(n))) % （10^9 + 7)

解法：

  1、暴力求 g(n) % (10 ^ 9 + 7) 的循环节  —— 222222224LL

  2、暴力求 g(n) % (222222224LL) 的循环节 —— 183120LL

  3、正推， 求出答案。

代码：

/*
 * =====================================================================================
 *
 *         Author:  KissBuaa.DS(AC)
 *        Company:  BUAA-ACMICPC-Group
 *
 * =====================================================================================
 */
/*
#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
#define LL long long
const LL modo = 222222224LL; //打表
LL a[3];
int main(){
    freopen("mengzhu.txt","w",stdout);
    a[0] = 0LL , a[1] = 1LL;
    LL i = 2LL;
    for( ; ; i= i + 1LL){
        a[ i % 3 ] = (a[ (i + 1) % 3] + 3LL * a[ (i + 2 ) % 3])%modo;
        if (a[ i % 3 ] == 1 && a[ (i+2) % 3] ==0) break;
    }
    cout<< i << endl;
}

222222224LL
183120LL
*/


#include <stdio.h>
#include <stdlib.h>
#include <string.h>
//#include <stdbool.h>
#include <math.h>
#define LL __int64
#define CLR(x) memset(x,0,sizeof(x))
#define typec LL
#define sqr(x) ((x)*(x))
#define abs(x) ((x)<0?(-(x)):(x))
#define min(a,b) ((a)<(b)?(a):(b))
#define max(a,b) ((a)>(b)?(a):(b))
#define PI acos(-1.0)
#define lowbit(x) ((x)&(-(x)))
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
#define inf 100000000
//For C++
#include <cctype>
#include <algorithm>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <bitset>
#include <list>
#include <iostream>
using namespace std;
const double eps=1e-10;
int dblcmp(typec d) {
    if (fabs(d)<eps)
       return 0;
    return (d>0)?1:-1;
}
LL modo;
const int maxn=3;
class Mat
{
public:
       LL Matn,Matm;
       typec a[maxn][maxn];
       Mat()
       {
            Matn=0;
            Matm=0;
            memset(a,0LL,sizeof(a));
       }
       void output();
       void init();
       void initI();
       Mat mul(const Mat &a);
       Mat power(const Mat&a,LL k);
};
void Mat::output()
{
     for (int i=0;i<Matn;++i)
     {
         for (int j=0;j<Matm;++j)
         {
             if (j!=0) printf(" ");
             printf("%I64d",a[i][j]);
         }
         printf("\n");
     }
}
void Mat::init()
{
    Matn=0;
    Matm=0;
    memset(a,0LL,sizeof(a));
}
Mat Mat::mul(const Mat &A)
{
    Mat c;
    c.init();
    c.Matn=Matn;
    c.Matm=A.Matm;
    for (int i=0;i<Matn;++i)
        for (int j=0;j<A.Matm;++j)
        {
            for (int k=0;k<Matm;++k)
            {
                c.a[i][j]=(c.a[i][j]+(a[i][k]*A.a[k][j])%modo)%modo;
            }
        }
    return c;
}
void Mat::initI()
{
     memset(a,0LL,sizeof(a));
     for (int i=0;i<Matn;++i) a[i][i]=1LL;
}
Mat Mat::power(const Mat& a,LL  k)
{
    Mat c=a,b;
    b.init();
    b.Matn=a.Matn;b.Matm=a.Matm;
    b.initI();
    while (k)
    {
         if (k & (1LL))
            b=b.mul(c);
         c=c.mul(c);
         k>>=1LL;
    }
    return b;
}
Mat  A ,B;
LL g(LL x){
    //Mat A;
    if (x == 0LL) return 0LL;
    if (x == 1LL) return 1LL;
    //A.init();
    //A.Matn = 1LL , A.Matm = 2LL;
    //A.a[0][0]=1LL,A.a[0][1]=0LL;
    //Mat B;
    B.init();
    B.Matn = 2LL , B.Matm = 2LL;
    B.a[0][0]= 3LL, B.a[1][0]=1LL;
    B.a[0][1] = 1LL; B.a[1][1]=0LL;
    B = B.power(B , x - 1LL);
    //A = A.mul(B);
    return B.a[0][0];
}
LL n;
void solve(){
    //printf("%I64d\n",g(42837));
    modo = 183120LL;
    LL res = g(n);
    //res = (res * inv(res , modo))% modo;
    //cout<<res<<endl;
    modo = 222222224LL;
    res = g(res);
    //cout<<res<<endl;
    //res = (res * inv(res , modo))% modo;
    modo =(1000000000LL + 7LL);
    res = g(res);
    //res = (res * inv(res-1LL , modo))% modo;
    //printf("%I64d\n",g(g(g(n))));
    printf("%I64d\n",res);
}
/*0
0
set<LL> has;
void solve(){
    has.clear();
    LL i;
    for (i = 1 ; ; ++i){
        LL res = g(i);
        if(has.find(res) != has.end()){
            cout<<i<<endl;
            return;
        }
        has.insert(res);
    }
}*/
int main(){
    //solve();
    while (~scanf("%I64d",&n)) solve();
}