杭电OJ1005题

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

问题分析：

根据f(n)的定义公式，f(n)的取值范围是[0,6].则组合（f(n-1),f(n-2))共有49种可能：(0,0),(0,1),……(6,6),f(n)肯定在这49种组合的49个值之内,而且在前49个值之后，f(n)的取值将照此循环，所以只要计算存储前49个值即可。

提交代码

#include<iostream>
using namespace std;

int main()
{
int arr[50];
int A,B,n;
arr[1]=1;
arr[2]=1;

while(cin>>A>>B>>n && (A|B|n))
{
for(int i=3;i<50;i++)
{
arr[i]= (arr[i-1]*A+arr[i-2]*B)%7;
}
cout<<arr[n%49]<<endl;
}
return 0;
}