HDU 1005 Number Sequence（矩阵）

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 177761    Accepted Submission(s): 44124

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

 

 

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这道题需要推一个类似于斐波那契矩阵的矩阵。

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#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int MAXN=1001;
inline void read(int &n){char c='+';bool flag=0;n=0;    
while(c<'0'||c>'9') c=='-'?flag=1,c=getchar():c=getchar();    
while(c>='0'&&c<='9') n=n*10+c-48,c=getchar();flag==1?n=-n:n=n;}
struct matrix
{
    int m[3][3];matrix(){memset(m,0,sizeof(m));}
};
matrix ma;
int limit=2;
const int mod=7;
matrix mul(matrix a,matrix b)
{
    matrix c;
    for(int k=0;k<limit;k++)
        for(int i=0;i<limit;i++)
            for(int j=0;j<limit;j++)
                c.m[i][j]=(c.m[i][j]+(a.m[i][k]*b.m[k][j]))%mod;
    return c;
}
matrix fast_martix_pow(matrix ma,int p)
{
    matrix bg;
    bg.m[0][0]=1;bg.m[1][1]=1;
    bg.m[0][1]=0;bg.m[1][0]=0;
    /*for(int i=0;i<limit;i++)
    {
        for(int j=0;j<limit;j++)
            cout<<bg.m[i][j]<<" ";
        cout<<endl;
    }*/
        
    while(p)
    {
        if(p&1)    bg=mul(bg,ma);
        ma=mul(ma,ma);
        p>>=1;
    }
    return bg;
}
int main()
{
    int a,b,n;
    while(scanf("%d%d%d",&a,&b,&n)&&(a!=0&&b!=0&&n!=0))
    {
        ma.m[0][0]=a;ma.m[0][1]=b;
        ma.m[1][0]=1;ma.m[1][1]=0;
        if(n<3)
        {
            printf("1\n");
            continue;
        }
        matrix ans=fast_martix_pow(ma,n-2);
        printf("%d\n",(ans.m[0][1]+ans.m[0][0])%mod);
    }
    return 0;
}