hdu3306 Another kind of Fibonacci(构造矩阵)

Another kind of Fibonacci

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1944    Accepted Submission(s): 747

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 <= 2 ³¹ – 1
X : 2<= X <= 2 ³¹– 1
Y : 2<= Y <= 2 ³¹ – 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  

/*
真心不好构造{S(n-2), a(n-1)^2,a(n-1)*a(n-2),a(n-2)^2}，刚好可以a(n)*a(n-1)=a(n-1)（X*a(n-1)+Y*a(n-2)）还有点坑，注意x,y提前取一下模，否则会溢出
加油！！！
Time:2015-4-14 21:22
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
const int mod=10007;
struct Matrix{
    LL mat[5][5];
    int n;
    void Init(int _n){
        n=_n;
        memset(mat,0,sizeof(mat));
    }
    Matrix operator *(const Matrix &b) const{
        Matrix ret;
        ret.Init(n);
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                for(int k=0;k<n;k++){
                    ret.mat[i][j]+=mat[i][k]*b.mat[k][j];
                    ret.mat[i][j]%=mod;
                }
            }
        }
        return ret;
    }
};
Matrix mat_pow(Matrix a,int k){
    Matrix ret;ret.Init(a.n);
    for(int i=0;i<ret.n;i++)ret.mat[i][i]=1;
    while(k>0){
        if(k&1) ret=ret*a;
        a=a*a;
        k>>=1;
    }
    return ret;
}
int main(){
    int n,x,y;
    Matrix trans,ret;
    while(scanf("%d%d%d",&n,&x,&y)!=EOF){
        trans.Init(4);ret.Init(4);
        x%=mod; y%=mod;
        trans.mat[0][0]=1; trans.mat[1][0]=1;
        trans.mat[1][1]=x*x%mod; trans.mat[2][1]=2*x*y%mod; trans.mat[3][1]=y*y%mod;
        trans.mat[1][2]=x; trans.mat[2][2]=y; trans.mat[1][3]=1;

        ret.mat[0][0]=1;//s0
        ret.mat[0][1]=1;//a1^2
        ret.mat[0][2]=1;//a1*a0
        ret.mat[0][3]=1;//a0^2
        trans=mat_pow(trans,n);
        ret=ret*trans;
        printf("%d\n",ret.mat[0][0]);
        /*
        for(int i=0;i<4;i++){
            for(int j=0;j<4;j++){
                printf("%d ",trans.mat[i][j]);
            }puts("");
        }
        */
    }
return 0;
}