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

 

仔细看一下数据范围，n的取值达到一亿。由于数据范围很大，因此这题不适合暴力。

那么这道类似斐波那契数列的题目该如何搞呢？很明显，f(n-1)和f(n-2)的取值只可能有0, 1, 2, 3, 4, 5, 6这七种情况。因此f(n-1)+f(n-2)的组合一共有7*7种情况，这说明什么呢？说明数列（以f3作为第一个数）出现循环最多在第50个数（最坏的情况，可由鸽巢原理得出）。

 

#include<cstdio>
int main()
{
    int a,b,n;
    while(scanf("%d%d%d",&a,&b,&n)!=EOF)
    {
        int ans[55]={0};
        int f1,f2,i;
        if(a==0&&b==0&&n==0)
            break;
        f1=f2=1;
        for(i=0;;i++)
        {
            ans[i]=(a*f2+b*f1)%7;
            f1=f2;
            f2=ans[i];
            if(i>=3&&ans[i-1]==ans[0]&&ans[i]==ans[1])
                break;
        }
        if(n==1||n==2)
            printf("1\n");
        else
            printf("%d\n",ans[(n-3)%(i-1)]);
    }
    return 0;
}