Number Sequence(HDU 1005 构造矩阵 )

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 147964    Accepted Submission(s): 35964

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).

 

 

Sample Input

1 1 3

1 2 1

0 0 0 0

 

Sample Output

2

5

 因为数据量比较大，可以打表找到循环规律，但这种类型的题发现都可以构造矩阵求解

f[n]　　　　　　=  a  b   *   f[n-1]

f[n-1]                  1  0　　  f[n-2]

 

 

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
int a,b,n;
int m[1000];
struct Matrix
{
    int mat[2][2];
}p;
Matrix mul(Matrix a,Matrix b)
{
    Matrix c;
    for(int i=0;i<2;i++)
    {
        for(int j=0;j<2;j++)
        {
            c.mat[i][j]=0;
            for(int k=0;k<2;k++)
                c.mat[i][j]=(c.mat[i][j]+a.mat[i][k]*b.mat[k][j])%7;
        }
    }
    return c;
}
Matrix mod_pow(Matrix x,int n)
{
    Matrix res;
    memset(res.mat,0,sizeof(res.mat));
    for(int i=0;i<2;i++)
        res.mat[i][i]=1;
    while(n)
    {
        if(n&1)
            res=mul(res,x);
        x=mul(x,x);
        n>>=1;
    }
    return res;
}
int main()
{
    freopen("in.txt","r",stdin);
    while(scanf("%d%d%d",&a,&b,&n)!=EOF)
    {
        if(a==0&&b==0&&n==0)
            break;
        if(n<2)
        {
            printf("1
");
            continue;
        }
        p.mat[0][0]=a;
        p.mat[1][0]=1;
        p.mat[0][1]=b;
        p.mat[1][1]=0;
        Matrix ans= mod_pow(p,n-2);
        printf("%d
",(ans.mat[0][0]+ans.mat[0][1])%7);
    }
}