杭电 1005

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 124649    Accepted Submission(s): 30286

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

 

分析：

因为n的取值范围太大，不优化直接计算肯定水不过；

因为A,B的值在运算时是固定不变的，所以肯定是一个周期数列；

f[i]和f[i-1]的取值只可能是0,1,2,3,4,5,6，所以A * f(n - 1) + B * f(n - 2)最多有7*7=49种可能的组合；

所以周期最大为49；

因为f[1]=f[2]=1,所以从f[3]开始遍历，只要出现f[i]和f[i-1]都为1的情况则说明进入了下一个周期；

那么周期就是i-2；

最后根据n和周期的值输出相应的数‘。

代码：

#include<iostream>
using namespace std;
int main()
{
    int a, b, n;
    while (cin >> a >> b >> n, a + b + n)
    {
        int f[55];
        f[2] = f[1] = 1;
        int zhouqi,i;
        for (i = 3;i <= 49;i++)
        {
            f[i] = (f[i - 1] * a + f[i - 2] * b) % 7;
            if (f[i] == f[i - 1] &&f[i]== 1)
                break;
        }
        zhouqi = i - 2;
        n = n%zhouqi;
        if (n == 0)
            cout << f[zhouqi] << endl;
        else
            cout << f[n] << endl;
    }
    return 0;
}