Number Sequence--hdu1005

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 128332    Accepted Submission(s): 31212

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

这个题其实应该可以考虑到有循环的，看了好多博客说直接能得出来循环节为49，但是我太笨，不明白！

我还是自己找吧！一切数据都是由f[1]=1和f[2]=1演化出来的，所以当再次出现时，一定是在循环了！

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 int f[];
 int main()
 {
     int i,j,a,b,n;
     f[]=;
     f[]=;
     while(scanf("%d%d%d",&a,&b,&n),a||b||n)
     {
         int t;
         for(i=;i<;i++)
         {
             f[i]=(a*f[i-]+b*f[i-])%;
             if(f[i-]==&&f[i]==)
             break;

         }
             i-=;
             n=n%i;
             if(n==)//当n=0时，说明正好是循环节的倍数，把它转化为一个循环的最后一个
             n=i;
             printf("%d\n",f[n]);
     }
     return ;
  }