ORIGIN

HDU-1005 Number Sequence

2019-12-19

Number Sequence

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
2
3

1 1 3
1 2 10
0 0 0

Sample Output

1
2

2
5

Analysis

This problem seems like a very huge problem, however, the answers have some certain regular pattern. For $f[n]$, it’s consisted of a basic number array. That means after a certain number of numbers, the numbers will be the same as begin. Like $1, 2, 5, 6, 1, 2, 5, 6, 1, 2, 5, 6 \cdots$. So just loop to find the recycle number.

Code

Not that if you don’t define boarder for i, it will be time limit exceeded. And the biggest recycled number is no bigger than 49.

#include<bits/stdc++.h>

using namespace std;

int f[100];
int main() {
	int a, b, n;
	while(cin >> a >> b >> n, a, b, n) {
		f[1] = f[2] = 1;
		int i = 3;
		for(; i < 50; i ++) {
			f[i] = (a * f[i - 1] + b * f[i - 2]) % 7;
			if(f[i - 1] == 1 and f[i] == 1) {
				break;
			}
		}
		f[0] = f[i - 2];
		cout << f[n % (i - 2)] << endl;
	}
	return 0;
}