HDU1588Gauss Fibonacci(矩阵)

Gauss Fibonacci

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2076    Accepted Submission(s): 898

Problem Description

Without expecting, Angel replied quickly.She says: "I'v heard that you'r a very clever boy. So if you wanna me be your GF, you should solve the problem called GF~. "
How good an opportunity that Gardon can not give up! The "Problem GF" told by Angel is actually "Gauss Fibonacci".
As we know ,Gauss is the famous mathematician who worked out the sum from 1 to 100 very quickly, and Fibonacci is the crazy man who invented some numbers.

Arithmetic progression:
g(i)=k*i+b;
We assume k and b are both non-nagetive integers.

Fibonacci Numbers:
f(0)=0
f(1)=1
f(n)=f(n-1)+f(n-2) (n>=2)

The Gauss Fibonacci problem is described as follows:
Given k,b,n ,calculate the sum of every f(g(i)) for 0<=i<n
The answer may be very large, so you should divide this answer by M and just output the remainder instead.

 

Input

The input contains serveral lines. For each line there are four non-nagetive integers: k,b,n,M
Each of them will not exceed 1,000,000,000.

 

Output

For each line input, out the value described above.

 

Sample Input

2 1 4 1002 0 4 100

 

Sample Output

2112
分析：
 列出前n项分别为f(b),f(k+b),f(2*k+b)......f(（n-1）*k+b)
结合斐波那契数列
设A为斐波那契数列的递推矩阵则原式课化为
A^b（E，A^k，A^2K，......A^(n-1)k）*(0,1);
设R=A^k 原式课化为 A^b（E，R^，R^2，......R^(n-1)）*(0,1)
 P=R,E ,p^n 右上角的子矩阵为 E，R^，R^2，......R^(n-1)
   0,E
剩下的事情就很简单了。
代码如下：
#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int N =4;int k,b,n,M;struct matrix{    long long m[4][4];};matrix I={   1,0,0,0,   0,1,0,0,   0,0,1,0,   0,0,0,1};matrix multi(matrix a,matrix b){    matrix c;    for(int i=0;i<N;i++)        for(int j=0;j<N;j++){                c.m[i][j]=0;            for(int k=0;k<N;k++)                c.m[i][j]+=(a.m[i][k]%M*b.m[k][j]%M)%M;            c.m[i][j]%=M;        }    return c;}matrix pow(matrix A,int  k){    matrix ans=I,p=A;    while(k){        if(k&1){            ans=multi(ans,p);            k--;        }        k>>=1;        p=multi(p,p);    }    return ans;}int main(){    while(cin>>k>>b>>n>>M){        matrix A={            0,1,0,0,            1,1,0,0,            0,0,0,0,            0,0,0,0        };        matrix B=pow(A,b);        matrix R=pow(A,k);        matrix W=R;        W.m[0][2]=1,W.m[1][3]=1,W.m[2][2]=1,W.m[3][3]=1;        matrix Q=pow(W,n);        matrix P;        for(int i=0;i<N;i++)            for(int j=0;j<N;j++)                P.m[i][j]=0;        P.m[0][0]=Q.m[0][2],P.m[0][1]=Q.m[0][3];        P.m[1][0]=Q.m[1][2],P.m[1][1]=Q.m[1][3];        matrix ans=multi(P,B);        cout<<(ans.m[0][1]+M)%M<<endl;        /*cout<<"*****B*****"<<endl;        for(int i=0;i<N;i++){            for(int j=0;j<N;j++)                cout<<B.m[i][j]<<" ";            cout<<endl;        }        cout<<"*****R*****"<<endl;        for(int i=0;i<N;i++){            for(int j=0;j<N;j++)                cout<<R.m[i][j]<<" ";            cout<<endl;        }        cout<<"*****Q*****"<<endl;        for(int i=0;i<N;i++){            for(int j=0;j<N;j++)                cout<<Q.m[i][j]<<" ";            cout<<endl;        }        cout<<"*****P*****"<<endl;        for(int i=0;i<N;i++){            for(int j=0;j<N;j++)                cout<<P.m[i][j]<<" ";            cout<<endl;        }*/    }    return 0;}