Another kind of Fibonacce(矩阵快速幂，HDU3306）

Another kind of Fibonacci

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2470    Accepted Submission(s): 987

Problem Description

As we all known , the Fibonacci series : F(0) = 1, F(1) = 1, F(N) = F(N - 1) + F(N - 2) (N >= 2).Now we define another kind of Fibonacci : A(0) = 1 , A(1) = 1 , A(N) = X * A(N - 1) + Y * A(N - 2) (N >= 2).And we want to Calculate S(N) , S(N) = A(0)² +A(1)²+……+A(n)².

 

Input

There are several test cases.
Each test case will contain three integers , N, X , Y .
N : 2<= N <= 2³¹ – 1
X : 2<= X <= 2³¹– 1
Y : 2<= Y <= 2³¹ – 1

 

Output

For each test case , output the answer of S(n).If the answer is too big , divide it by 10007 and give me the reminder.

 

Sample Input

2 1 1
3 2 3

 

Sample Output

6
196

 

Author

wyb

 

Source

HDOJ Monthly Contest – 2010.02.06

 

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注意：用cin>>输入会时间超限，改用scanf即可。

#include <iostream>
#include <cstdio>

using namespace std;

typedef long long ll;

#define mod 10007

ll n;
ll x;
ll y;

struct matrix
{
 ll v[4][4];
};

matrix mul(matrix m1, matrix m2)
{
 matrix c;
 for (int i = 0; i<4; i++)
  for (int j = 0; j<4; j++)
  {
   c.v[i][j] = 0;
   for (int k = 0; k<4; k++)
    c.v[i][j] += (m1.v[i][k] * m2.v[k][j]) % mod;
   c.v[i][j] %= mod;
  }
 return c;
}

matrix qupow(matrix A)
{
 n--;
 matrix res = A, m = A;
 while (n)
 {
  if (n & 1)
   res = mul(res, m);
  n >>= 1;
  m = mul(m, m);
 }
 return res;
}

void solve(matrix A)
{
 matrix temp = qupow(A);
 ll ans = 0;
 for (int i = 0; i<4; i++)
  ans += temp.v[i][0]%mod;
    ans%=mod;
    printf("%lld\n",ans);
 //cout << ans << endl;
}

int main()
{
 matrix A;
 ll a, b, c;
 while (~scanf("%lld%lld%lld",&n,&x,&y))
 {
        a = x*x%mod;
  b = y*y%mod;
  c = x*y * 2%mod;
  //matrix A = {  1,0,0,0 ,1,a,1,x,0,b,0,0 0,c,0,y };
  A.v[0][0]=1;
  for(int i=1;i<4;i++)
            A.v[0][i]=0;
        A.v[1][0]=1;
        A.v[1][1]=a%mod;
        A.v[1][2]=1;
        A.v[1][3]=x%mod;
        A.v[2][0]=0;
        A.v[2][1]=b%mod;
        A.v[2][2]=0;
        A.v[2][3]=0;
        A.v[3][0]=0;
        A.v[3][1]=c%mod;
        A.v[3][2]=0;
        A.v[3][3]=y%mod;
  /*for(int i=0;i<4;i++)
            for(int j=0;j<4;j++)
            cin>>A.v[i][j];*/
  solve(A);
 }
 return 0;
}