YT14-HDU-求斐波那契数列

最新推荐文章于 2022-07-19 08:58:36 发布

原创最新推荐文章于 2022-07-19 08:58:36 发布 · 875 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#c++ #编程 #iostream #namespace #数据

HDU 专栏收录该内容

48 篇文章

订阅专栏

本文介绍了一种解决特定数列问题的方法，该数列定义为f(1)=1,f(2)=1,f(n)=(A*f(n-1)+B*f(n-2))mod7。通过分析给出的代码示例，探讨了如何高效地计算f(n)，并提供了完整的C++实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

代码如下：

#include <iostream>
using namespace std;
int num[10000];
int main()
{
    int A,B,n;
    num[1] = num[2] = 1;
    while(cin>>A>>B>>n, A !=0|| B!=0 || n!=0)
    {
        int i;
        for(i=3; i<10000 ; i++)
        {
            num[i] = (A*num[i-1] + B*num[i-2]) % 7;
            if(num[i] == 1 &&num[i-1] == 1)
                break;
        }
        n = n % (i-2);
        num[0] = num[i-2];
        cout << num[n] << endl;
    }
    return 0;
}

运行结果：