Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 83849    Accepted Submission(s): 19863

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

 

 

Source

 

 

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JGShining

#include
      
       int main(){    int a,b,n;    while(scanf("%d %d %d",&a,&b,&n)&&(a!=0|b!=0||n!=0))    {        int k=3,f1=1,f2=1,f3;        n=n%48;        if(n==1||n==2)            printf("%d\n",n);        else        {            while(k<=n)            {                f3=(b*f1+a*f2)%7;                f1=f2;                f2=f3;                k++;                   }            printf("%d\n",f3);        }            }    return 0;    }
      

这题有意思的地方就是如果你直接用递归或循环求就会超时。经过百遍的观察，我们可以发现对7求余结果也就是0 1 2 3 4 5 6，所以应该是7*7个一循环，但是，结果是48个一循环，为什么会是48呢，我现在也不太清楚，我是先没做过处理，然后一个一个找的规律，得出48循环，谁知道为什么麻烦告诉我一声。