hdu4291 A Short problem 矩阵快速幂 求循环节----成都网络赛

 

A Short problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 300    Accepted Submission(s): 110

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

   
   
   
   

    
    
    
    0
1
2
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    0
1
42837
   
   
   
   

 

Source

2012 ACM/ICPC Asia Regional Chengdu Online

 

Recommend

liuyiding

 

主要是求循环节。其实不难.

 

 

//求循环节
#include<iostream>
#include<stdio.h>
using namespace std;
typedef long long LL;
const LLMOD = 1000000007LL;
int main() 
{
    LL f0 = 0, f1 = 1, temp = -1;
    for (LL i = 1;; i++) 
    {
        temp = (3 * f1 + f0) % MOD;
        f0 = f1;
        f1 = temp;
        if (f0 == 0 && f1 == 1) 
        {
            printf("%I64d\n", i);
            break;
        }
    }
    return 0;
}

//const LL MOD = 1000000007LL;
//222222224

//const LL MOD = 222222224LL;
//183120



//A题代码
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<stdio.h>
#include<math.h>
#define ll __int64
using namespace std;
const int MAX = 2;
ll mm;
typedef struct
{
ll  m[MAX][MAX];
} Matrix;
Matrix P =
{
3,1,
1,0
};
Matrix I =
{
1,0,
0,1
};
Matrix matrixmul(Matrix a,Matrix b)
{
int i,j,k;
Matrix c;
for (i = 0 ; i < MAX; i++)
for (j = 0; j < MAX;j++)
{
c.m[i][j] = 0;
for (k = 0; k < MAX; k++)
c.m[i][j] += (a.m[i][k] * b.m[k][j]%mm+mm)%mm;
c.m[i][j] = (c.m[i][j]%mm+mm)%mm;
}
return c;
}

Matrix quickpow(ll n)
{
Matrix m = P, b = I;
while (n >= 1)
{
if (n & 1)
b = matrixmul(b,m);
n = n >> 1;
m = matrixmul(m,m);
}
return b;
}

int main()
{
    ll n;
    while(scanf("%I64d",&n)!=EOF)
    {

       if(n==0)
       {
           cout<<0<<endl;
           continue;
       }
       else if(n==1)
       {
           cout<<1<<endl;
           continue;
       }
       mm=183120;
       Matrix g=quickpow2(n-1);
       ll yy=g.m[0][0];

       mm=222222224LL;
       if(yy!=0&&yy!=1)
       {
        g=quickpow2(yy-1);
        yy=g.m[0][0];
       }


       mm=1000000007LL;
       if(yy!=0&&yy!=1)
       {
           g=quickpow2(yy-1);
           yy=g.m[0][0]%mm;
       }
       printf("%I64d\n",yy);
    }
}