Number Sequence(快速幂矩阵）

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=1005

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 131753    Accepted Submission(s): 31988

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

 

题意：简单的快速幂矩阵，加一个f(n-1) = f(n-1)即可

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
struct Mtr{
    int a , b , c , d ;
    Mtr operator * (const Mtr m) const
    {
        Mtr res ;
        res.a = (a*m.a%7 + b*m.c%7)%7;
        res.b = (a*m.b%7+b*m.d%7)%7;
        res.c = (c*m.a%7+d*m.c%7)%7;
        res.d = (c*m.b%7+d*m.d%7)%7;
        return res;
    }
};
int cal(Mtr a , int n)
{
    Mtr c;
    c.a = c.d = 1;
    c.b = c.c = 0;
    while(n>0)
    {
        if(n&1)
            c = c*a;
        a = (a*a);
        n/=2;
    }
    return (c.a+c.b)%7;
}
int main()
{
    int a , b , n;
    while(~scanf("%d%d%d",&a,&b,&n)&&(a!=0||b!=0||n!=0))
    {
        Mtr tm;
        tm.a = a;
        tm.b = b;
        tm.c = 1;
        tm.d = 0;
        int tt = cal(tm,n-2);
        printf("%d
",tt);
    }
    return 0;
}