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 31 2 100 0 0

 

Sample Output

25

 

#include <iostream>using namespace std;int modValue(int aA,int bB,int nN);int main(){   int A,B,n;   while(cin>>A>>B>>n&&A!=0&&B!=0&&n!=0         &&A>=1&&A<=1000&&B>=1&&B<=1000         &&n>=1&&n<=100000000)   {       int value,t;       t=n%49;        value=modValue(A,B,t);        cout<<value<<endl;   }    return 0;}int modValue(int aA,int bB,int nN){    int x;    if(nN==1||nN==2)    x=1;    else        x=(aA*modValue(aA,bB,nN-1)+bB*modValue(aA,bB,nN-2))%7;        return x;}

虽然看了很多大神解释的循环周期为49，但还是有点一直半解，自己也有自己的理解方式，只是看起来自己理解的比较简单而已