Number Sequence&&http://acm.hdu.edu.cn/showproblem.php?pid=1005-优快云博客

本文链接：https://blog.youkuaiyun.com/smallacmer/article/details/7178342

本文介绍了一道关于数列周期的问题，通过模7运算确定数列周期，利用周期性快速计算第n项的值。输入包括参数A、B及n，输出为数列的第n项。

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Number Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 48952 Accepted Submission(s): 10923

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
一看这题的时间要求就知道常规的方法肯定不行了，，于是把目标锁定在mod7上，想了好打一会mod7这说明f（n）的取值仅在0-6之间，，那就是说如果给定一组a和b后F[N]共有49种取法所以周期肯定小于49，因此这题找到周期就ok了
AC代码
#include<iostream>
using namespace std;
int f[50];
int main()
{ int a,b,m,i;
  while(cin>>a>>b>>m&&a&&b&&m)
  { f[1]=f[2]=1;
    for(i=3;i<=50;++i)
     {
     f[i]=((a*f[i-1])%7+(b*f[i-2])%7)%7;
     if(f[i]==f[i-1]&&f[i-1]==1) break;
     }
     i=i-2;
     m%=i;
     if(m) cout<<f[m]<<endl;
     else  cout<<f[i]<<endl;
     }return 0;
    }