HDU1005 Number Sequence（思维&规律）

problem:
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

AC代码：

#include
int main()
{
    int a,b,n,i,s[49];
    while(~scanf("%d%d%d",&a,&b,&n),a||b||n)
    {
        s[1]=1;
        s[2]=1;
        for(i=3;i<49;i++)
        {
            s[i]=(a*s[i-1]+b*s[i-2])%7;
        }
        if(n%49==0) printf("%d",s[1]);
        else printf("%d",s[n%49]);
        printf("\n");
    }
    return 0;
}

思路：因为每一个f(n)都是对7取余后得来的，所以f(n-1)与f(n-2)各有0,1,2,3,4,5,6这7种情况，每一个f(n)由f(n-1)与f(n-2)得出(a,b的值输入后即是固定值)，所以n总共只有49种结果即49个结果为一个循环，所以需用数组储存49个结果，当输入n时通过对49取余的结果找出答案。（思路不完全，待续）