HDU 1005：Number Sequence【规律】

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 138216    Accepted Submission(s): 33500

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

 对于公式 f[n] = A * f[n-1] + B * f[n-2]; 后者只有7 * 7 = 49 种可能，为什么这么说，因为对于f[n-1] 或者 f[n-2] 的取值只有 0,1,2,3,4,5,6 这7个数，A，B又是固定的，所以就只有49种可能值了。由该关系式得知每一项只与前两项发生关系，所以当连续的两项在前面出现过循环节出现了，注意循环节并不一定会是开始的 1，1 。 又因为一组测试数据中f[n]只有49中可能的答案，最坏的情况是所有的情况都遇到了，那么那也会在50次运算中产生循环节。找到循环节后，就可以轻松解决了。

AC-code:

#include<cstdio>int f[60];int main(){int a,b,i;long long n;f[1]=f[2]=1;while(scanf("%d%d%lld",&a,&b,&n),a|b|n){for(i=3;i<=49;i++)f[i]=(f[i-1]*a%7+f[i-2]*b%7)%7;printf("%d\n",f[n%48]);}return 0;}