Number Sequence

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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

Author

CHEN, Shunbao

分析：本题题意很明显，但是给的范围是100000000；范围较大，而对于求余问题，很容易想到循环节；

当分析几组数据后，有出现1的，所以不能用一个1来判断，很容易想到连续两个数都为1时，找到循环节；

关于如何找循环节：雀鸽原理：求关于7的余数，其余数必小于7，所以连续的两个数有7*7=49种；用一循环即可；

代码：

#include<stdio.h>#include<string.h>#define max 100000000int f[max];int main(){int n,i,a,b,m;while(~scanf("%d%d%d",&a,&b,&n)&&!(a==0&&b==0&&n==0)){//memset(f,0,sizeof(f));  不能有这一语句，有了会超内存，毕竟max=100000000,较大 f[1]=f[2]=1;for(i=3;i<51;i++)  {f[i]=(a*f[i-1]+b*f[i-2])%7;//printf("%d ",f[i]);if((f[i]==f[i-1])&&(f[i]==1))    {m=i-2;break;   }  }  //printf("%d\n",m);  n=n%m;  if(n==0)  //当n==0时也是一种情况；   n=m;   printf("%d\n",f[n]);}return 0;}

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