hdu 1588 Gauss Fibonacci(矩阵乘法)

题目：​​http://acm.hdu.edu.cn/showproblem.php?pid=1588​​

Gauss Fibonacci

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2851    Accepted Submission(s): 1181

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

 

Author

DYGG

 

Source

​​HDU “Valentines Day” Open Programming Contest 2007-02-14​​

 

推导：

所以计算过程中涉及到了两个不同规模的矩阵。当然也可以用一种大矩阵表示两种不同的小矩阵，在矩阵计算中控制好边界即可：

#include <iostream>
#include <cstdio>
using namespace std;
typedef long long LL;
struct matrie1{
    LL m[2][2];
};
struct matrie2{
    LL m[4][4];
};
matrie1 A={
    0,1,
    1,1,
};
matrie1 AI={
    1,0,
    0,1
};
matrie2 B={
0,0,1,0,
0,0,0,1,
0,0,1,0,
0,0,0,1
};
matrie2 BI={
1,0,0,0,
0,1,0,0,
0,0,1,0,
0,0,0,1
};
int k,b,n,mod;
matrie1 multi1(matrie1 a,matrie1 b){
    matrie1 q;
    for(int i=0;i<2;i++){
        for(int j=0;j<2;j++){
            q.m[i][j]=0;
            for(int k=0;k<2;k++){
                q.m[i][j]+=a.m[i][k]*b.m[k][j]%mod;
            }
            q.m[i][j]%=mod;
        }
    }
    return q;
}
matrie1 pow1(int k){
    matrie1 q=AI,tmp=A;
    while(k){
        if(k&1)q=multi1(q,tmp);
        tmp=multi1(tmp,tmp);
        k>>=1;
    }
    return q;
}
matrie2 multi2(matrie2 a,matrie2 b){
    matrie2 q;
    for(int i=0;i<4;i++){
        for(int j=0;j<4;j++){
            q.m[i][j]=0;
            for(int k=0;k<4;k++){
                q.m[i][j]+=a.m[i][k]*b.m[k][j]%mod;
            }
            q.m[i][j]%=mod;
        }
    }
    return q;
}
matrie2 pow2(int k){
    matrie2 q=BI,tmp=B;
    while(k){
        if(k&1)q=multi2(q,tmp);
        tmp=multi2(tmp,tmp);
        k>>=1;
    }
    return q;
}
int main()
{
    //freopen("cin.txt","r",stdin);
    while(~(scanf("%d %d %d %d",&k,&b,&n,&mod))){
         matrie1 R=pow1(k);
         for(int i=0;i<2;i++){
             for(int j=0;j<2;j++)B.m[i][j]=R.m[i][j];
         }
         matrie2 BN=pow2(n);
         matrie1 B01;
         for(int i=0;i<2;i++){
             for(int j=2;j<4;j++)B01.m[i][j-2]=BN.m[i][j];
         }
         matrie1 Ab=pow1(b);
         matrie1 res=multi1(Ab,B01);
         printf("%d\n",res.m[0][1]%mod);
    }
    return 0;
}