hdu_1005_Number Sequence_201310222120

Number Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 86547 Accepted Submission(s): 20560

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

Author

CHEN, Shunbao

Source

ZJCPC2004

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又是一道给出了运算公式的数学，凡是没有优化的话，超时超内存等等是避免不了的了。(错了好多次就是因为这，百度了下才明白)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 <stdio.h>
 2 #include <string.h>
 3 
 4 int main()
 5 {    
 6     int a,b,n;
 7     while(scanf("%d %d %d",&a,&b,&n),a||b||n)
 8     {
 9         int i,xh=0;
10         int f[100];
11         memset(f,0,sizeof(f));
12         f[1]=f[2]=1;
13         for(i=3;i<100;i++)
14         {
15             f[i]=(a*f[i-1]+b*f[i-2])%7;
16             if(f[i]==1&&f[i-1]==1)
17             break;
18         }
19         xh=i-2;
20         n=n%xh;
21         if(n==0)
22         n=xh;
23         //for(i=1;i<60;i++)
24         //printf("%d ",f[i]);
25         printf("%d
",f[n]);
26     }
27     return 0;
28 }

网上简洁做法：

http://www.189works.com/article-19050-1.html