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 + ;
 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, , sizeof(a));
         a[] = , a[] = ;
         bool flag = ;
         int start, len;
         for(int i = ; ; i++)
         {
             a[i] = n * a[i - ] + m * a[i - ];
             a[i] %= ;
             for(int j = ; j <= i - ; j++)
             {
                 if(a[j] == a[i - ] && a[j + ] == a[i])
                 {
                     flag = ;
                     start = j;
                     len = i - j - ;
                 }
                 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 ;
 }