HDOJ 1005

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 3

1 2 10

0 0 0

 

Sample Output

2

5

 

#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <string>

using namespace std;

typedef long long ll;

ll f(ll n,ll A,ll B){
    if(n == 1||n == 2)return 1;
    return  (A*f(n-1,A,B)+B*f(n-2,A,B))%7;
}

int main(){
    ll n,A,B;
    while(cin>>A>>B>>n){
        if(n==0&&A==0&&B==0)break;
        cout<<f(n%49,A,B)<<endl;
    }
    return 0;
}