1005

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

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这题不能直接按公式用递归来求，因为n最大可以达到100,000,000，会栈溢出
所以要找规律
前两个等于1，所以后面如果有两个连着的1出现，那就是出现周期了.

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int main(){    int a,b,n;    while (scanf ("%d%d%d",&a,&b,&n) && (a+b+n))    {        int i,j;        int f[100]={0,1,1};        for (i=3; i<100; i++)        {            f[i] = (a*f[i-1] + b*f[i-2]) % 7;       //如果有两个连着 =1，则后面的全部和前面相同，即出现了周期        //这时就没必要再进行下去了，跳出循环, i-2为周期               if (f[i] == 1 && f[i-1] == 1)                break;        }        n = n%(i-2);        f[0] = f[i-2];  /* 把n对周期求模，当n %(i-2)==0时,此时本来应该取f[i-2]的，所以把f[0]=f[i-2]*/        printf ("%d\n",f[n]);    }    return 0;}