hdu1588之经典矩阵乘法

 

Gauss Fibonacci

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1706    Accepted Submission(s): 741

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

题目要求求出f(g(i))的总和,i是0~n-1

 

代码中详细思路+注释

 

/*f(g(i))=f(k*i+b)

令f[n]=A;//A是矩阵,A的某个元素是F[n]

若i=0~n-1,则sum(f(k*i+b))

=A^b+A^(k+b)+A^(2k+b)+A^(3k+b)+...+A^((n-1)k+b)

=A^b+A^b(A^k+A^2k+A^3k+A^4k+...+A^(n-1)k)

将A^k看成一个新的矩阵B,则原式:

=A^b+A^b(B^1+B^2+B^3+...+B^(n-1));//A^b,A^k用矩阵快速幂求出,括号中的用二分矩阵可求

所谓二分矩阵:A^1+A^2+A^3+A^4+A^5+A^6=(A^1+A^2+A^3)+A^3(A^1+A^2+A^3)

*/

#include<iostream>

#include<cstdio>

#include<cstdlib>

#include<cstring>

#include<string>

#include<queue>

#include<algorithm>

#include<map>

#include<iomanip>

#define INF 99999999

using namespace std;



const int MAX=2;

__int64 array[MAX][MAX],sum[MAX][MAX],temp[MAX][MAX],ans[MAX][MAX];

//array相当于A,sum记录每次幂乘后的矩阵,temp是临时矩阵,ans是B^1+B^2+B^3+...+B^n 



void MatrixInit(__int64 a[MAX][MAX],bool flag){//初始化矩阵 

	for(int i=0;i<MAX;++i){

		for(int j=0;j<MAX;++j){

			if(flag)a[i][j]=array[i][j];//a=A

			else a[i][j]=(i == j);//a=1

		}

	}

}



void MatrixAdd(__int64 a[MAX][MAX],__int64 b[MAX][MAX],int &mod){//矩阵相加 

	for(int i=0;i<MAX;++i){//a=a+b

		for(int j=0;j<MAX;++j){

			a[i][j]=(a[i][j]+b[i][j])%mod;

		}

	}

}



void MatrixMult(__int64 a[MAX][MAX],__int64 b[MAX][MAX],int &mod){//矩阵相乘 

	__int64 c[MAX][MAX]={0};

	for(int i=0;i<MAX;++i){//a=a*b

		for(int j=0;j<MAX;++j){

			for(int k=0;k<MAX;++k){

				c[i][j]+=a[i][k]*b[k][j];

			}

		}

	}

	for(int i=0;i<MAX;++i){

		for(int j=0;j<MAX;++j)a[i][j]=c[i][j]%mod;

	}

}



void MatrixPow(int k,int &mod){//矩阵幂乘,sum=A^k 

	MatrixInit(sum,0);//sum=1

	MatrixInit(temp,1);//temp=A

	while(k){

		if(k&1)MatrixMult(sum,temp,mod);

		MatrixMult(temp,temp,mod);

		k>>=1;

	}

}



void MatrixSum(int k,int &mod){//矩阵求和 

	if(k == 1){MatrixInit(ans,1);return;}

	MatrixSum(k/2,mod);

	MatrixPow((k+1)/2,mod);

	if(k&1){//k为奇数则A+(A+A^m)*(A+A^2+A^3...),m=(k+1)/2 

		MatrixInit(temp,1);//temp=A

		MatrixAdd(sum,temp,mod);//sum=A+A^m

		MatrixMult(ans,sum,mod);//ans=sum*ans

		MatrixAdd(ans,temp,mod);//ans=A+ans

	}

	else{//k为偶数则(1+A^m)*(A+A^2+A^3...),m=(k+1)/2 

		MatrixInit(temp,0);//temp=1

		MatrixAdd(temp,sum,mod);//temp=1+A^m

		MatrixMult(ans,temp,mod);//ans=ans*temp;

	}

}



int main(){

	int k,b,n,m;

	while(scanf("%d%d%d%d",&k,&b,&n,&m)!=EOF){

		array[0][0]=array[0][1]=array[1][0]=1;

		array[1][1]=0;

		MatrixPow(k,m);//求A^k

		MatrixInit(array,0);

		MatrixMult(array,sum,m);//将array构造成B,即A^k

		MatrixSum(n-1,m);//求矩阵和

		array[0][0]=array[0][1]=array[1][0]=1;

		array[1][1]=0;

		MatrixPow(b,m);//求A^b;

		MatrixMult(ans,sum,m);//求A^b*ans

		MatrixAdd(ans,sum,m);//求A^b+A^b+ans

		printf("%I64d\n",ans[1][0]); 

	}

	return 0;

}