杭电ACM 二数学求模

时间：2015-02-12 22:50:53 阅读：178 评论：0 收藏：0 [点我收藏+]

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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

/*  
*Copyright (c)2014,烟台大学计算机与控制工程学院  
*All rights reserved.  
*文件名称：test.cpp  
*作    者：冷基栋  
*完成日期：2015年2月12日  
*版 本 号：v1.0  
*问题描述：数学求模
*/
#include <iostream>
using namespace std;
int arr[100];
int main()
{
    std::ios::sync_with_stdio(false);
    int A,B,n;
    arr[1] = arr[2] = 1;
    while(cin>>A>>B>>n,A||B||n)
    {
        int i;
        for(i=3; i<100 ; i++)
        {
            arr[i]=(A*arr[i-1]+B*arr[i-2])%7;
            if(arr[i]==1&&arr[i-1]==1)
                break;
        }
        n=n%(i-2);
        arr[0]=arr[i-2];
        cout << arr[n] << endl;
    }
    return 0;
}

运行结果：

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