HDOJ_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

思路：递归化成迭代会超时，需要空间换时间，题目的mod7代表f(n-1)和f(n-2)都是0-6的变化范围故周期为49

#include <iostream>using namespace std;int main(){    int a, b, n;    while (cin>>a>>b>>n)    {        if (a == 0 && b ==0 && n ==0)        {            break;        }        int buf[54] = { 0,1,1 };        int z = 1;        for (int i = 3; i < 54; ++i)        {            buf[i] = (a*buf[i-1] + b*buf[i-2]) % 7;            if (i > 5)            {                if (buf[i-1] == buf[3] && buf[i] == buf[4])                {                    z = i - 4;                    break;                }            }        }        if (n > 2)        {            cout << buf[(n - 3) % z + 3] << endl;        }        else            cout << 1 << endl;    }    return 0;}