找规律：（0，0），（0，1）......（6，6）；每次a和b的值是一样的。找规律，每49一次循环。

 

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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 AC代码：

#include 
      
       #include 
       
        using namespace std;int main(void){    freopen("in.txt","r",stdin);    int a,b;    long long n;    while(scanf("%d%d%lld",&a,&b,&n)!=EOF&&(a||b||n))    {        int x[100];        x[1]=1;        x[2]=1;        for(int i=3;i<=49;i++)            x[i]=((a*x[i-1])%7 + (b*x[i-2])%7)%7;                 printf("%d\n",x[n%49]);    }        fclose(stdin);    return 0;}