* Fibonacci sequence (take surplus)

Problem K
Time limit:1000 MS Memory limit:32 MB 64bit IO Format:%i64d
submitted:225 accepted:51
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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 is 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 in a single line (1 <= A, b <=, 1 <= n <= 100,000,000). Three zeros signal the end of the input and this test case are not a is processed.

Output

For each test case, print the value of f (n) in a single line.

Sample Input

1 1 3
1 2 10
0 0 0

Sample Output

2
5

Since the value of MOD7,F (n) can only be taken in 0-6, and A and B are not changed, it is bound to loop, and the loop section will not exceed 7*7=49;

#include <stdio.h>
int main ()
{
    long  a,b,n,i,s,q,p;
    while (scanf ("%i64d%i64d%i64d", &a,&b,&n)!=eof)
    {
        if (a==0&&b==0&&n==0)
            break;
        P=1;
        Q=1;
        if (n==1| | n==2)
            printf ("1\n");
        else
        {
            n=n%49;
            for (i=3; i<=n; i++)
            {
                s= (a*q+b*p)%7;
                p=q;
                q=s;
            }
            printf ("%i64d\n", s);
        }

    }
    return 0;
}