HDU 1005 Number Sequence 矩阵快速幂

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 236241    Accepted Submission(s): 60070

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

 

解题思路：

就是一个简单的矩阵快速幂的模板题，推出他的传递矩阵是

{ A，B，

  1， 0}；

然后再套用矩阵快速幂的模板即可，注意当询问为n的时候，要求幂的次数是n-2次

 

代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#define mem(a) memset(a,0,sizeof(a))
#define forn(i,n) for(int i=0;i<n;++i)
#define for1(i,n) for(int i=1;i<=n;++i)
#define IO std::ios::sync_with_stdio(false); std::cin.tie(0)
using namespace std;
typedef long long ll;
const int maxn=30;
const int mod=7;
const int inf=0x3f3f3f3f;

int n,m,k,t,x,y;
struct Matrix
{
    int val[2][2];
};

Matrix E=
{1,0,
 0,1
};

Matrix Multiply(Matrix a,Matrix b)
{
    Matrix ans=E;
    forn(i,2)
    forn(j,2)
    {
        int temp=0;
        forn(k,2)
        temp=(temp+a.val[i][k]*b.val[k][j]%mod)%mod;
        ans.val[i][j]=temp;
    }
    return ans;
}

int quick_pow(Matrix a,int n)
{
    Matrix res=E;
    while(n>0)
    {
        if(n&1)
        res=Multiply(res,a);
        a=Multiply(a,a);
        n>>=1;
    }
    int ans=0;
    ans=(res.val[0][0]+res.val[0][1])%mod;
    return ans;
}

int main(){
    IO;
    while(cin>>x>>y>>n)
    {
        if(x==y&&n==x&&x==0) break;
        Matrix t=
        {
            x,y,
            1,0
        };
        int ans=(quick_pow(t,n-2))%mod;
        cout<<ans<<endl;
    }
    return 0;
}