HDU1005

Number Sequence

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 31 2 100 0 0

 

Sample Output

25
//HDU1005 基础题//Number Sequence//2017.05.23 by wyj#includeusing namespace std;int f[50];int main(){int A, B, n;int i, j, t;f[1] = f[2] = 1;while (cin >> A >> B >> n, A | B | n){int begin, end, flag = 0;for (i = 3;(i <= n) && (!flag);i++){f[i] = (A * f[i - 1] + B * f[i - 2]) % 7;for (j = 2;j < i;j++){if (f[i] == f[j] && f[i - 1] == f[j - 1]){begin = j - 1;//循环起点end = i - 1;//循环终点flag = 1;//超出循环结标志break;}}}if (flag)//如果超出第一个循环t = f[begin + (n - end) % (end - begin)];else//不超出第一个循环t = f[n];cout << t << endl;}return 0;}