HDU OJ 1005 Number Sequence

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 178988    Accepted Submission(s): 44485

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 31 2 100 0 0

 

Sample Output

25

 

Author

CHEN, Shunbao

 

Source

ZJCPC2004

 

Recommend

JGShining   |   We have carefully selected several similar problems for you:  1008 1021 1019 1009 1012 

直接用递归或者循环肯定超时，爆内存。因为  1 <= n <= 100,000,000

所以 关键在于  mod 7.

要找其循环周期。

如果是一个f()的话 循环周期是7

现在是两个f()所以 循环周期是49；

代码：

#include <iostream>  using namespace std;  int arr[50];  int main()  {      int n,a,b;      arr[1]=arr[2]=1;      while(cin>>a>>b>>n)      {          if(n==0) break;          for(int i=3; i<=50; i++)          {              arr[i]=(a*arr[i-1]+b*arr[i-2])%7;          }          cout<<arr[n%49]<<endl;      }      return 0;  }  

Submit