HDU 1005 Number Sequence-优快云博客

本文链接：https://blog.youkuaiyun.com/AleiChen/article/details/50505532

探讨如何通过发现数列循环节简化f(n)计算，利用矩阵快速幂解决特定形式递推数列的高效计算问题。

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我就在想到底有没有循环节，最后我决定，本题目是有循环节的，因为只要出现两个相邻的数值相同，那么他就会重置f(n)的值进入循环，至于是不是1，是不能确定的。如果出现5和5相邻，是不是就从5开始循环了，这也是确定的。这题目真是有点乱。原解释矩阵快速幂。不过还搞不懂，加油。

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

代码：

#include<stdio.h>
#include<iostream>

using namespace std;

int a[10005];
int A, B, n;
int main()
{
    a[1] = a[2] = 1;
    while(scanf("%d%d%d", &A, &B, &n), A || B || n)
    {
        int i;
        for(i=3; i<10000; i++)
        {
           a[i] = (A*a[i-1] + B*a[i-2]) % 7;
           if(a[i] == 1 && a[i-1] == 1)
            break;
        }
        n = n % (i-2);
        a[0] = a[i-2];
        cout<<a[n]<<endl;
    }
    return 0;
}