HDUOJ 1005 Number Sequence

直接用递归模拟会超内存，考虑到对7取余，A,B为定值，所以可以进行打表

 1 #include <iostream>
 2 #include <stdio.h>
 3 using namespace std;
 4 int main()
 5 {
 6     int a, b, n, i, t;
 7     int f[50];
 8     f[1] = f[2] = 1;
 9     while((3 == scanf("%d%d%d", &a, &b, &n)), a || b || n)
10     {
11         for(i = 3; i <= 49; ++i)
12         {
13             f[i] = (a * f[i-1] + b * f[i-2]) % 7;
14             if(1 == f[i] && 1 == f[i-1]) break;       //cycle found
15         }
16         t = i - 2;         //cycle
17         f[0] = f[t];       //将周期的最后一个放到0位置。实现f(n%t)
18         printf("%d\n", f[n%t]);
19     }
20     return 0;
21 }