[ACMcoder] Number Sequence

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

解题思路

利用数组的循环规律解题，周期为i-2。

实现代码

#include <iostream>using namespace std;int main(){    int A, B, n;    int f[53];    while (cin>>A>>B>>n)    {        if (A == 0 && B == 0 && n == 0)        {            break;        }        f[1] = 1; f[2] = 1;        int i, j, z;        for (i = 3; i <= 52; i++)        {            f[i] = (A * f[i-1] + B * f[i-2]) % 7;            if (f[i] == 1 && f[i-1] == 1)            {                break;            }        }        i -= 2;        n %= i;        if (n == 0)        {            cout<<f[i]<<endl;        }        else        {            cout<<f[n]<<endl;        }    }};