zoj 2105 Number Sequence

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原文链接：http://www.cnblogs.com/forever4444/archive/2009/05/09/1453166.html

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#c/c++

本文解析了ZOJ Problem Set中的2105号问题——Number Sequence。该问题是关于递推数列的计算，在给定特定递推公式的情况下，需要求解数列中第n项的值。文章提供了详细的算法思路及C++实现代码。

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ZOJ Problem Set - 2105

Number Sequence

Time Limit: 1 Second Memory Limit: 32768 KB

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

Author: CHEN, Shunbao

Source: Zhejiang Provincial Programming Contest 2004
Submit Status

// 1385524 2008-03-29 22:34:43 Accepted  2105 C++ 0 832 Wpl
#include < iostream >
using namespace std;
int main()
{
     int a,b,d;
     long n;
     int arry[ 49 ];
    arry[ 1 ] = 1 ;
    arry[ 2 ] = 1 ;
    cin >> a >> b >> n;
     while (a != 0 && b != 0 && n != 0 )
    {
         for (d = 3 ;d <= 48 ;d ++ )
            arry[d] = (arry[d - 1 ] * a % 7 + arry[d - 2 ] * b % 7 ) % 7 ;
        cout << arry[(n - 1 ) % 48 + 1 ] << endl;
        cin >> a >> b >> n;
    }
}

转载于:https://www.cnblogs.com/forever4444/archive/2009/05/09/1453166.html