HDU-1005 Number Sequence

题目：

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

解析：

1. 本题是一个找规律的题，对于%7来说保证了f(n)最多有7钟，f(n+1)也是7种，所以f(n+2)最多49种。所以49之后就是规律。下图是我打表出来的图。第一张为1 1 {1~50}，循环周期为16，第二张是1,2，{1~50}，循环周期是6

2. 总结打表的规律是，当出现两个连续的1时，说明此时进入下一个循环。

#include
int main()
{
    int a, b, n;
    int s[101] = { 0 ,1,1};
    while (~scanf("%d%d%d", &a, &b, &n) && (a + b + n))
    {
        int m = 0,i;
            for (i = 3; i < 100; i++)
        {
            s[i] = (a*s[i - 1] + b*s[i - 2])%7;
            if (s[i] == 1 && s[i - 1] == 1)
                break;
        }
            m = i - 2;//周期
            n %= m;
            if(n!=0)
            printf("%d\n", s[n]);
            else
                printf("%d\n", s[i-2]);
    }
    return 0;
}