HDU1588Gauss Fibonacci(矩陣)

bigbigship發表於2014-07-26
矩陣

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 4 100
2 0 4 100





 



Sample Output




21
12
分析：
 列出前n項分別為f(b),f(k+b),f(2*k+b)......f(（n-1）*k+b)
結合斐波那契數列
設A為斐波那契數列的遞推矩陣則原式課化為
A^b（E，A^k，A^2K，......A^(n-1)k）*(0,1);
設R=A^k 原式課化為 A^b（E，R^，R^2，......R^(n-1)）*(0,1)
 P=R,E ,p^n 右上角的子矩陣為 E，R^，R^2，......R^(n-1)
   0,E
剩下的事情就很簡單了。
程式碼如下：
#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;
}






            




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