hdu1005 Number Sequence---找循环节

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=1005
题目大意：

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).
思路：

由于模7，循环节一定在49以内，所以计算前49位，找到循环节，直接输出a[n]即可。

start为循环节开始的下标，len为循环节长度，计算a[n]时，如果 n <= start，直接输出a[n],如果大于start，输出a[start + (n - start) % len]；

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<set>
#include<cmath>
using namespace std;
const int maxn = 1e4 + 10;
typedef long long ll;
int T, n, m;
int a[maxn];
int main()
{
    int x;
    while(cin >> n >> m >> x && (n + m + x))
    {
        memset(a, 0, sizeof(a));
        a[1] = 1, a[2] = 1;
        bool flag = 0;
        int start, len;
        for(int i = 3; ; i++)
        {
            a[i] = n * a[i - 1] + m * a[i - 2];
            a[i] %= 7;
            for(int j = 1; j <= i - 2; j++)
            {
                if(a[j] == a[i - 1] && a[j + 1] == a[i])
                {
                    flag = 1;
                    start = j;
                    len = i - j - 1;
                }
                if(flag)break;
            }
            if(flag)break;
        }/*
        for(int i = 1; i < start; i++)cout<<a[i]<<" ";
        for(int i = start; i <= start + len; i++)cout<<a[i]<<" ";
        cout<<endl;
        cout<<start<<" "<<len<<endl;*/
        if(x <= start)cout<<a[x]<<endl;
        else cout<<a[start + (x - start) % len]<<endl;
    }
    return 0;
}