HDU 1588（数论，构造二分矩阵+求幂运算+二分求和）

admin管理员组

文章数量:1026989

Gauss Fibonacci

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 437    Accepted Submission(s): 204

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
  
  
  
  
   
   /*******************************************************************************************
	首先我们看g=k*i+b; 是一个等差数列

	如果能推出f(g)这个函数也是一个等差或者等比数列，就可以得出一个公式

	f(k*i+b)  0<=i<n

	建立一个二分矩阵A=[1 1,1 0]
	
	f(b)=A^b  

	则:

	f(b)=A^b

	f(k+b)=A^k+b

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

	我们就可以得出一个等比数列:

	首项：A^b
	
	公比：A^k

	项数：n

	(res是单位矩阵)

	运用等比数列求和公式得出：sum=A^b*(res+A^k+(A^k)^2+(A^k)^3+...+(A^k)^(n-1))

	需要注意的一点是：当b=0的情况
***************************************************************************************/

/***************************************************************************************

第一次这种思路做测试数据都过了，可以TLE了~ 看来主要在
for(i=1;i<n;i++)
	temp=Martix_Add(temp,er_fun(q,i));	
这里超时了，要想办法优化下。

----------想了好久，决定用二分求和------------
例如:A+A^2+A^3+A^4+A^5+A^6=(A+A^2+A^3)+A^3*(A+A^2+A^3)  用递归写

***************************************************************************************/
#include <iostream>
using namespace std;
#define ll __int64
#define N 2
struct Mat
{
	ll martix[N][N];
};

Mat tp,res,A,q,a1,temp;
ll mod,k,b,n;

Mat Martix_Add(Mat a,Mat b)
{
	int i,j;
	Mat c;
	for (i=0;i<N;i++)
	{
		for (j=0;j<N;j++)
		{
			c.martix[i][j]=(a.martix[i][j]+b.martix[i][j])%mod;
		}
	}
	return c;
}
Mat Martix_Mul(Mat a,Mat b)
{
	int i,j,l;
	Mat c;
	for (i=0;i<N;i++)
	{
		for (j=0;j<N;j++)
		{
			c.martix[i][j]=0;
			for (l=0;l<N;l++)
			{
				c.martix[i][j]+=(a.martix[i][l]*b.martix[l][j])%mod;
				c.martix[i][j]%=mod;
			}
		}
	}
	return c;
}

Mat er_fun(Mat e,ll x)  //求矩阵e^x
{
	tp=e;
	e=res;
	while(x)
	{
		if(x&1)
			e=Martix_Mul(e,tp);
		tp=Martix_Mul(tp,tp);
		x>>=1;
	}
	return e;
}

Mat Solve(Mat a,ll p)
{
	if(p==1) return a;
	else if(p&1) return Martix_Add(er_fun(a,p),Solve(a,p-1));
	else return Martix_Mul(Solve(a,p>>1),Martix_Add(er_fun(a,p>>1),res));
}

int main()
{
	int i,j;
	for(i=0;i<N;i++)
	{
		for (j=0;j<N;j++)
		{
			if(i==j)
				res.martix[i][j]=1;
			else
				res.martix[i][j]=0;
		}
	}
	A.martix[0][0]=1;
	A.martix[0][1]=1;
	A.martix[1][0]=1;
	A.martix[1][1]=0;
	while (scanf("%I64d%I64d%I64d%I64d",&k,&b,&n,&mod)!=EOF)
	{
		q=A;
		q=er_fun(q,k);
		temp=res;
		/*for(i=1;i<n;i++)
			temp=Martix_Add(temp,er_fun(q,i));	*/
		temp=Martix_Add(temp,Solve(q,n-1));
		if(b!=0)
		{
			a1=er_fun(A,b);
			temp=Martix_Mul(a1,temp);
		}
		printf("%I64d/n",temp.martix[0][1]%mod);
	}
	return 0;
} 
  
  

Gauss Fibonacci

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 437    Accepted Submission(s): 204

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
  
  
  
  
   
   /*******************************************************************************************
	首先我们看g=k*i+b; 是一个等差数列

	如果能推出f(g)这个函数也是一个等差或者等比数列，就可以得出一个公式

	f(k*i+b)  0<=i<n

	建立一个二分矩阵A=[1 1,1 0]
	
	f(b)=A^b  

	则:

	f(b)=A^b

	f(k+b)=A^k+b

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

	我们就可以得出一个等比数列:

	首项：A^b
	
	公比：A^k

	项数：n

	(res是单位矩阵)

	运用等比数列求和公式得出：sum=A^b*(res+A^k+(A^k)^2+(A^k)^3+...+(A^k)^(n-1))

	需要注意的一点是：当b=0的情况
***************************************************************************************/

/***************************************************************************************

第一次这种思路做测试数据都过了，可以TLE了~ 看来主要在
for(i=1;i<n;i++)
	temp=Martix_Add(temp,er_fun(q,i));	
这里超时了，要想办法优化下。

----------想了好久，决定用二分求和------------
例如:A+A^2+A^3+A^4+A^5+A^6=(A+A^2+A^3)+A^3*(A+A^2+A^3)  用递归写

***************************************************************************************/
#include <iostream>
using namespace std;
#define ll __int64
#define N 2
struct Mat
{
	ll martix[N][N];
};

Mat tp,res,A,q,a1,temp;
ll mod,k,b,n;

Mat Martix_Add(Mat a,Mat b)
{
	int i,j;
	Mat c;
	for (i=0;i<N;i++)
	{
		for (j=0;j<N;j++)
		{
			c.martix[i][j]=(a.martix[i][j]+b.martix[i][j])%mod;
		}
	}
	return c;
}
Mat Martix_Mul(Mat a,Mat b)
{
	int i,j,l;
	Mat c;
	for (i=0;i<N;i++)
	{
		for (j=0;j<N;j++)
		{
			c.martix[i][j]=0;
			for (l=0;l<N;l++)
			{
				c.martix[i][j]+=(a.martix[i][l]*b.martix[l][j])%mod;
				c.martix[i][j]%=mod;
			}
		}
	}
	return c;
}

Mat er_fun(Mat e,ll x)  //求矩阵e^x
{
	tp=e;
	e=res;
	while(x)
	{
		if(x&1)
			e=Martix_Mul(e,tp);
		tp=Martix_Mul(tp,tp);
		x>>=1;
	}
	return e;
}

Mat Solve(Mat a,ll p)
{
	if(p==1) return a;
	else if(p&1) return Martix_Add(er_fun(a,p),Solve(a,p-1));
	else return Martix_Mul(Solve(a,p>>1),Martix_Add(er_fun(a,p>>1),res));
}

int main()
{
	int i,j;
	for(i=0;i<N;i++)
	{
		for (j=0;j<N;j++)
		{
			if(i==j)
				res.martix[i][j]=1;
			else
				res.martix[i][j]=0;
		}
	}
	A.martix[0][0]=1;
	A.martix[0][1]=1;
	A.martix[1][0]=1;
	A.martix[1][1]=0;
	while (scanf("%I64d%I64d%I64d%I64d",&k,&b,&n,&mod)!=EOF)
	{
		q=A;
		q=er_fun(q,k);
		temp=res;
		/*for(i=1;i<n;i++)
			temp=Martix_Add(temp,er_fun(q,i));	*/
		temp=Martix_Add(temp,Solve(q,n-1));
		if(b!=0)
		{
			a1=er_fun(A,b);
			temp=Martix_Mul(a1,temp);
		}
		printf("%I64d/n",temp.martix[0][1]%mod);
	}
	return 0;
} 
  
  

本文标签： 数论矩阵HDU