HDU 1005 Number Sequence

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
#include <iostream>
using namespace std;
int f[1000];//定义在外面范围大一些;
int main()
{
    int a,b,n;
    while(cin>>a>>b>>n)
    {
        if(a==0&&b==0&&n==0)
        break;
        f[1]=1;
        f[2]=1;
        int flag;
        for(int i=3;i<=1000;i++)
        {
            f[i]=(a*f[i-1]+b*f[i-2])%7;
            if(f[i]==f[i-1])//对于每一个数组,前两个元素大小都是1,都是相等的,所以可以作为判断是否循环的条件;
            {
                flag=i-2;//如果满足这个条件的话,i-1是找到的下一个循环节的起始位置,那么i-2就是上一个循环的结束;
            }
        }
        cout<<f[n%flag]<<endl;//flag就是循环周期;
    }
    return 0;
}
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