HDU1005Number Sequence(找规律)

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 152357    Accepted Submission(s): 37087

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

思路：打表找规律，a,b固定，mod7后只有0，1，2，3，4，5，6这几位数字，所以最长的周期是49，，也就是说第50个会是第一个，不排除在50前就循环

 #include<bits/stdc++.h>
 using namespace std;
 #define LL long long
 LL n,a,b;
 int main(){
     while(cin>>a>>b>>n){
         if(a==b&&b==&&n==b)
         return ;
         else
         {
             int aa[];
             aa[]=aa[]=;
             int i;
             for(i=;i<=;i++){
                 aa[i]=((a*aa[i-]+b*aa[i-]))%;
                 if(aa[i]==&&aa[i-]==)
                 break;
             }
             n=n%(i-);
             if(n==)
             cout<<aa[i-]<<endl;
             else
             cout<<aa[n]<<endl;
         }
     }
 }