HDU 1005

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

网上看到一句话，没有优化超时超内存是避免不了的。。。

关键是找周期吧！

#include<iostream>

using namespace std;

int main()
{
    int n,a,b;
    int rec[50];
    rec[1]=rec[2]=1;
    while(cin>>a>>b>>n,a||b||n)
    {
        int start,end,flag=0;
        for(int i=3;i<=n && !flag;i++)
        {
            rec[i]=(a*rec[i-1]+b*rec[i-2])%7;
            for(int j=2;j<i;j++)
            {
                if(rec[i]==rec[j] && rec[i-1]==rec[j-1])
                {
                    flag=1;start=j;end=i;
                    break;
                }
            }
        }
        if(flag) cout<<rec[start+(n-end)%(end-start)]<<endl;
        else cout<<rec[n]<<endl;
    }
    return 0;
}