[HDU2855]Fibonacci Check-up

题目：Fibonacci Check-up

链接：http://acm.hdu.edu.cn/showproblem.php?pid=2855

分析：

1）二项式展开：$(x+1)^n = sum^n_{k=0}{C^k_n * x^k}$

2）Fibonacci数列可以写为：$ left[ egin{array}{cc} 0 & 1 \ 1 & 1 end{array} ight]^n$的左下角项。

3）构造矩阵$ T = Fib+E = left[ egin{array}{cc} 0 & 1 \ 1 & 1 end{array} ight] + left[ egin{array}{cc} 1 & 0 \ 0 & 1 end{array} ight] = left[ egin{array}{cc} 1 & 1 \ 1 & 2 end{array} ight]$。

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
typedef long long LL;
int MOD;
struct Matrix{
    LL a[2][2];
    void init(int f){
        memset(a,0,sizeof a);
        if(f==-1)return;
        for(int i=0;i<2;++i)a[i][i]=1;
    }
};
Matrix operator*(Matrix& A,Matrix& B){
    Matrix C;C.init(-1);
    for(int i=0;i<2;++i)
    for(int j=0;j<2;++j)
        for(int k=0;k<2;++k){
            C.a[i][j]+=A.a[i][k]*B.a[k][j];
            C.a[i][j]%=MOD;
        }
    return C;
}
Matrix operator^(Matrix A,int n){
    Matrix Rt;Rt.init(0);
    for(;n;n>>=1){
        if(n&1)Rt=Rt*A;
        A=A*A;
    }
    return Rt;
}
int main(){
    int n,Case;scanf("%d",&Case);
    Matrix A,T;
    T.a[0][0]=1;T.a[0][1]=1;
    T.a[1][0]=1;T.a[1][1]=2;

    for(;Case--;){
        scanf("%d%d",&n,&MOD);
        A=T^n;
        LL ans=A.a[1][0];
        printf("%lld
",ans%MOD);
    }

    return 0;
}
        

4)$sum^n_{k=0}{C^k_n * f(k)} = f(2*n) $

5)证明：$ sum^n_{k=0}{C^k_n * f(k)} $

= $ sum^n_{k=0}{ C^k_n * { [ { ( frac{1+sqrt{5}}{2} )}^k - { ( frac{1-sqrt{5}}{2} )}^k }] } $

= $ sum^n_{k=0}{ C^k_n * {( frac{1+sqrt{5}}{2} )}^k } - sum^n_{k=0}{ C^k_n * { ( frac{1-sqrt{5}}{2} )}^k } $

= $ { ( frac{1+sqrt{5}}{2} + 1 ) }^k $ - $ { ( frac{1-sqrt{5}}{2} + 1 ) }^k $

= $ { ( frac{3+sqrt{5}}{2} ) }^k $ - $ { ( frac{3-sqrt{5}}{2} ) }^k $ 

= $ { ( frac{6+2*sqrt{5}}{4} ) }^k $ - $ { ( frac{6-2*sqrt{5}}{4} ) }^k $

= $ { ( frac{1+sqrt{5}}{2} ) }^{2k} $ - $ { ( frac{1-sqrt{5}}{2} ) }^{2k} $

= $ f(2*k) $

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
typedef long long LL;
int MOD;
struct Matrix{
    LL a[2][2];
    void init(int f){
        memset(a,0,sizeof a);
        if(f==-1)return;
        for(int i=0;i<2;++i)a[i][i]=1;
    }
};
Matrix operator*(Matrix& A,Matrix& B){
    Matrix C;C.init(-1);
    for(int i=0;i<2;++i)
    for(int j=0;j<2;++j)
        for(int k=0;k<2;++k){
            C.a[i][j]+=A.a[i][k]*B.a[k][j];
            C.a[i][j]%=MOD;
        }
    return C;
}
Matrix operator^(Matrix A,int n){
    Matrix Rt;Rt.init(0);
    for(;n;n>>=1){
        if(n&1)Rt=Rt*A;
        A=A*A;
    }
    return Rt;
}
int main(){
    int n,Case;scanf("%d",&Case);
    Matrix A,T;
    T.a[0][0]=0;T.a[0][1]=1;
    T.a[1][0]=1;T.a[1][1]=1;
    for(;Case--;){
        scanf("%d%d",&n,&MOD);
        A=T^(n+n);
        LL ans=A.a[1][0];
        printf("%lld
",ans%MOD);
    }

    return 0;
}