原题链接

Number Sequence

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

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

解题报告：

一开始做这题就是想暴力求解，所以如期的TLE了。然后换个思路，直接算出了通项公式，但是通项公式有两项都是带根号的，最后不能很好的转换为int，放弃！再接着，瞥了下discuss里面的讨论，得到的关键字是周期，其实也很容易想到，毕竟是mod 7，很快就应该有周期，所以只需要找到周期就好。注意！！！但是由于f(1)=f(2)=1是初始项，并不一定满足通项的结构，所以应该从f(3)开始匹配找周期！更详细的解释请移步：[ACM_HDU_1005]Number
Sequence

当然，这题还有另外的解法，详见 【HDU1005】Number Sequence（矩阵乘法）

#include <iostream>
using namespace std;
int a,b,n;
int A[100];
int main()
{
    while(cin>>a>>b>>n){
        if(!a && !b && !n) break;
        if(n==1 || n==2) cout<<1<<endl;
        else{
            A[1]=A[2]=1;int cyc=n;
            for(int i=3;i<=100;++i){
                A[i]=(a*A[i-1]+b*A[i-2])%7;
                if(i>5 &&  A[i]==A[4] && A[i-1]==A[3]){cyc=i-4;break;}
            }

            //输出
            cout<<A[(n-3)%cyc+3]<<endl;
        }
    }
    return 0;
}