HDU-1005 Number Sequence
Number Sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 43278 Accepted Submission(s): 9449
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).
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
又是一道给出了运算公式的数学，涛神还是没有说错，凡是没有优化的话，超时超内存等等是避免不了的了。这题很显然是一个找规律的题目，也就是该题的求解中是存在循环节的。
对于公式 f[n] = A * f[n-1] + B * f[n-2]; 后者只有7 * 7 = 49 种可能，为什么这么说，因为对于f[n-1] 或者 f[n-2] 的取值只有 0,1,2,3,4,5,6 这7个数，A，B又是固定的，所以就只有49种可能值了。由该关系式得知每一项只与前两项发生关系，所以当连续的两项在前面出现过循环节出现了，注意循环节并不一定会是开始的 1，1 。 又因为一组测试数据中f[n]只有49中可能的答案，最坏的情况是所有的情况都遇到了，那么那也会在50次运算中产生循环节。找到循环节后，就可以轻松解决了。
代码如下：
1 #include <cstring>
2 #include <cstdio>
3 #include <cstdlib>
4 using namespace std;
5
6 int rec[60];
7
8 int main()
9 {
10 int a, b, n;
11 rec[0] = rec[1] = rec[2] = 1;
12 while( scanf( "%d %d %d", &a, &b, &n ), a | b | n )
13 {
14 int beg, end, flag = 0;
15 for( int i = 3; i <= n && !flag; ++i )
16 {
17 rec[i] = ( a * rec[i-1] + b * rec[i-2] ) % 7;
18 for( int j = 2; j <= i - 1; ++j )
19 {
20 if( rec[i] == rec[j] && rec[i-1] == rec[j-1] )
21 {
22 beg = j, end = i;
23 flag = 1;
24 break;
25 }
26 }
27 }
28 if( flag )
29 {
30 printf( "%d\n", rec[beg+(n-end)%(end-beg)] );
31 }
32 else
33 printf( "%d\n", rec[n] );
34 }
35 return 0;
36 }
