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 31 2 100 0 0

 

Sample Output

25

该题给出f1，f2，输入A，B和N，让你求Fn的值(其中有规律，即Fn的值是有循环的，但不一定是从第一个开始的)

代码实现：

 #include<stdio.h>    int f[100000005];  int main()  {      int a,b,n,i,j;    f[1]=1;f[2]=1;      while(scanf("%d%d%d",&a,&b,&n))      {          int s=0;//记录周期长度         if(a==0&&b==0&&n==0) break;          for(i=3;i<=n;i++)          {              f[i]=(a*f[i-1]+b*f[i-2])%7;//利用前两个已知的值，便可求出后面的所有值//              for(j=2;j<i;j++)              if(f[i-1]==f[j-1]&&f[i]==f[j])            {                  s=i-j;//求出周期以后，便可推出后面所有的值，因此直接break//                  break;              }              if(s>0) break;          }          if(s>0){    f[n]=f[(n-j)%s+j];          }          printf("%d\n",f[n]);       }      return 0;  } 

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