hdu 1005 Number Sequence 矩阵连乘

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 37463    Accepted Submission(s): 7885

Problem Description

A number sequence is defined as follows:

f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.

Given A, B, and n, you are to calculate the value of f(n).

 

 

Input

The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed.

 

 

Output

For each test case, print the value of f(n) on a single line.

 

 

Sample Input

   
   
   
   

    
    
    
    1 1 3
1 2 10
0 0 0
   
   
   
   

 

 

Sample Output

   
   
   
   

    
    
    
    2
5
   
   
   
   

 

 

Author

CHEN, Shunbao

 

 

Source

ZJCPC2004

 

 

Recommend

JGShining

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=3;
int mod=7;
class Matrix
{
    int n,m;//矩阵的row and colum
    int num[maxn][maxn];//从1开始
    public:
        Matrix(){memset(num,0,sizeof(num));}
        Matrix(const Matrix &a) {n=a.n;m=a.m; memcpy(num,a.num,sizeof(a.num));}
        Matrix(int x,int y):n(x),m(y){}
        Matrix & operator =(const Matrix & a);
        friend Matrix operator + (const Matrix &a,const Matrix &b);
        friend Matrix operator - (const Matrix &a,const Matrix &b);
        friend Matrix operator * (const Matrix &a,const Matrix &b);
        friend Matrix operator % (const Matrix &a,int n);
        friend ostream & operator <<(ostream & cout,const Matrix & a);
        friend istream & operator >>(istream & cin,Matrix & a);
        void Modefy_n_m(int x,int y,int v);
        void out();//只在此题中有用，其他题目可删去
};
Matrix & Matrix:: operator =(const Matrix &a)
{
    for(int i=1;i<=a.n;i++) for(int j=1;j<=a.m;j++) num[i][j]=a.num[i][j];

    return *this;
}
istream & operator >>(istream & cin,Matrix & a)
{
    for(int i=1;i<=a.n;i++)
    {
        for(int j=1;j<=a.m;j++)
        {
            cin>>a.num[i][j];
        }
    }
    return cin;
}
ostream & operator <<(ostream & cout,const Matrix & a)
{
    for(int i=1;i<=a.n;i++)
    {
        for(int j=1;j<=a.m;j++)
        {
            cout<<a.num[i][j]<<"   ";
        }
        cout<<endl;
    }
    return cout;
}
Matrix operator % (const Matrix &a,int n)
{
    Matrix t;
    for(int i=1;i<=a.n;i++)
    {
        for(int j=1;j<=a.m;j++)
        {
            t.num[i][j]=a.num[i][j]%n;
        }
    }
    return t;
}
Matrix operator * (const Matrix &a,const Matrix &b)
{
    Matrix t(a.n,b.m);
    for(int i=1;i<=a.n;i++)
    {
        for(int j=1;j<=b.m;j++)
        {
            int cnt=0;
            for(int k=1;k<=a.m;k++)
            {
                cnt+=a.num[i][k]*b.num[k][j];
                cnt%=mod;//这是此题需要，别的题目可删去
            }
            t.num[i][j]=cnt;
        }
    }
    return t;
}
Matrix operator + (const Matrix &a,const Matrix &b)
{
    Matrix t;
    for(int i=1;i<=a.n;i++) for(int j=1;j<=a.m;j++) t.num[i][j]=a.num[i][j]+b.num[i][j];
    return t;
}
Matrix operator - (const Matrix &a,const Matrix &b)
{
    Matrix t;
    for(int i=1;i<=a.n;i++) for(int j=1;j<=a.m;j++) t.num[i][j]=a.num[i][j]-b.num[i][j];
    return t;
}
void Matrix:: Modefy_n_m(int x,int y,int v)
{
    num[x][y]=v;
}
void Matrix::out()
{
    printf("%d/n",num[1][2]);
}
Matrix E(2,2);

Matrix bin(Matrix cnt,int n)//cnt
{
    if(n==0) return E;
    if(n==1) return cnt;
    Matrix temp=bin(cnt,n/2);
    temp=temp*temp;
    if(n&1)
    {
        temp=temp*cnt;
        return temp;
    }
    else
    {
        return temp;
    }
}
int main()
{
    E.Modefy_n_m(1,1,1);E.Modefy_n_m(1,2,0);
    E.Modefy_n_m(2,1,0);E.Modefy_n_m(2,2,1);

    Matrix pre(1,2);
    pre.Modefy_n_m(1,1,1);pre.Modefy_n_m(1,2,1);
    int a,b,n;
    while(scanf("%d%d%d",&b,&a,&n)==3&&n)
    {
        if(n<3) {printf("1/n");continue;}
        n-=2;
        Matrix cnt(2,2);
        cnt.Modefy_n_m(1,1,0);cnt.Modefy_n_m(1,2,a);
        cnt.Modefy_n_m(2,1,1);cnt.Modefy_n_m(2,2,b);
        Matrix temp=bin(cnt,n);
        Matrix ans(1,2);
        ans=pre*temp;
        ans.out();
    }
    return 0;
}