HDU1005

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

这题就是找规律
一个递推函数考5个未知量确定一个值
其中先抛去f(n-1),f(n-2）来看
abn都是知道的
然后结果模7
说明f只可能是0到6
那么f(n-1),f(n-2)只有49种组合方式
因此代码如下

#include<iostream>
#include<algorithm>
#include<string>
using namespace std;
int f[81];
int main()
{
    int a, b, n;
    while (cin >> a >> b >> n)
    {
        if (a == 0 && b == 0 && n == 0)break;
        f[1] = 1;
        f[2] = 1;
        for (int v = 3;v <= 80;v++)f[v] = (f[v - 1] * a + f[v - 2] * b) % 7;
        int br = 0;
        int st, ed;
        for (int g = 1;g <= 78;g++)
        {
            for (int u = g + 1;u <= 79;u++)
            {
                if (f[u] == f[g] && f[u + 1] == f[g + 1])
                {
                    st = g + 2;
                    ed = u + 2;
                    br = 1;
                    break;
                }
            }
            if (br == 1)break;
        }
        ed = ed - st;
        st = st - 1;
        if (n < 80)cout << f[n] << endl;
        else
        {
            f[st - 1] = f[st + ed];
            cout << f[(n - st ) % ed + st] << endl;
        }
    }
    return 0;
}