1005 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 3
1 2 10
0 0 0

 

Sample Output

2
5

 

 #include<iostream>
 #define N 10000
 using namespace std;
 int main()
 {
     int A,B,n;
     cin>>A>>B>>n;
     n=n%;
     while(!(A==&&B==&&n==))
     {
         long long int a[N];
         a[]=;
         a[]=;
         if(n==||n==)
             cout<<<<endl;
         else
         {
             for(int i=;i<=n;i++)
             {
                 a[i]=(A*a[i-]+B*a[i-])%;
             }
             cout<<a[n]<<endl;
         }
         cin>>A>>B>>n;
         n=n%;
     }
     return ;
 }

由于f(n)是由前两个数字组合产生，那么只要有两个数字组合相同的情况发生就一定一会产生循环！

两个数字的组合的最大可能值为7x7=49，因此只要在调用迭代方法中限制n的在0~48就可以了