HDU1005 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 3
1 2 10
0 0 0
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    2
5
   
   
   
   

 

Author

CHEN, Shunbao

 

Source

ZJCPC2004

 

Recommend

JGShining   |   We have carefully selected several similar problems for you:   1002  1009  1108  1071  1006 

思路：这是一道找规律的题，结果共有49种可能。因为根据初步的计算，f[n]的取值只有：0，1，2，3，4，5，6，这七种可能，由于A和B的取值是确定的，所以共有7*7=49种可能。总结下来就是49为边界的循环。

代码如下：

#include<stdio.h>
int f[51]; int main() { int a,b; int n; int i;    
    f[1]=1;
    f[2]=1; while(scanf("%d%d%d",&a,&b,&n)!=EOF) { if(a==0&&b==0&&n==0) break; for(i=3;i<=49;i++) {
            f[i]=(a*f[i-1]+b*f[i-2])%7; }
        printf("%d\n",f[n%49]); } return 0; }