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==0&&n==b)
        return 0;
        else
        {
            int aa[100];
            aa[1]=aa[2]=1;
            int i;
            for(i=3;i<=50;i++){
                aa[i]=((a*aa[i-1]+b*aa[i-2]))%7;
                if(aa[i]==1&&aa[i-1]==1)
                break;
            }
            n=n%(i-2);
            if(n==0)
            cout<<aa[i-2]<<endl;
            else
            cout<<aa[n]<<endl; 
        }
    }
}