Number Sequence 找规律

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本文介绍了一种通过寻找规律来高效计算特定数列值的方法。该数列定义为f(1)=1,f(2)=1,f(n)=(A*f(n-1)+B*f(n-2))mod7。文章提供了一个C语言实现的例子，展示了如何通过预计算有限的状态来快速得出大数值n对应的数列值。

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Number Sequence
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 145337    Accepted Submission(s): 35307


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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来源： http://acm.hdu.edu.cn/showproblem.php?pid=1005

//找规律做法
#include <stdio.h>           // f(n) +7x=A*f(n-1) +B*f(n-2) // 2A7==49 f(n-1),f(n-2)排列   一共49情况
long long  Value[49]={0,1,1};

int A,B;

int Cal(long long n)

{

    for(int i=3;i<=48;i++)

        Value[i] = (A*Value[i-1]+B*Value[i-2])%7;

    return 1;

}

int main()

{



    long long n;

    while(~scanf("%d %d %lld",&A,&B,&n)&&(A||B||n)&&Cal(n))

        printf("%lld\n",Value[n%48==0?48:n%48]);

    return 0;

}