杭电ACM1005

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 153662    Accepted Submission(s): 37490

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

Author

CHEN, Shunbao

 #include <stdio.h>

 int main()
 {
     int n,a,b,f[],i,j,flag,begin,end;
     f[]=;f[]=;
     while(scanf("%d%d%d",&a,&b,&n),a|b|n)
     {
         flag=;
         for(i=;i<=n&&flag;i++)
         {
             f[i]=(a*f[i-]+b*f[i-])%;
             for(j=;j<=i-;j++)
             {
                 if(f[i]==f[j]&&f[i-]==f[j-])
                 {
                     begin=j;
                     end=i;
                     flag=;
                     break;
                 }
             }
         }
         if(!flag)
             printf("%d\n",f[begin+(n-end)%(end-begin)]);
         else
             printf("%d\n",f[n]);
     }
     return ;
 }

f[n]的值只有0到6共7种，因此必然会随着n增加，出现f[n]=f[i].f[n-1]=f[i-1]的情况，此时便是循环节开始了，之后的计算只需要知道n到循环起始位的距离就能按规律得出f[n]。而f[n]为0~6，表达式中两个f（n）产生49种可能。因此随着n增大到2+49=51必然出现循环，我们也只须定义int f[51]即可，只定义到50便是错误。

由此可见，关键在找到循环节起始位。