HDU--1005

题目介绍

1. 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).

2. 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.

3. Output

   For each test case, print the value of f(n) on a single line.

4. Sample Input

   1	1	3
   1	2	10
   0	0	0

5. Sample Output

   2

   5

题目分析

由题目中可以看出 f(n) 的取值范围是 0,1,2,3,4,5,6, 总共是7个数，因此 f(n-1) 和 f(n-2) 总共有 7*7 = 49种组合方法，于是对于一组给定的 A/B 的值， f(n)一共有 49 种可能的值，也就是说不管给定的 n 多大， 这组数中总共也只有 49 种数，因此可以推想这是一个循环，循环数为 49 。 经过验证，确实是这样的，因此可以相应的编写代码。

代码

#include <iostream>

using namespace std;

int main()
{
    int A, B, n;
    int fn[49];
    fn[0] = 1;
    fn[1] = 1;
    while(cin>>A>>B>>n)
    {   
        if(A==0 && B==0 && n==0)
        {
            break;
        }
        for(int i=2; i<49; i++)
        {
            fn[i] = (A * fn[i-1] + B * fn[i-2]) % 7;
        }
        cout<<fn[(n-1)%49]<<endl;
    }   
    return 0;
}