HDU 1005 Number Sequence 找规律

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).

[align=left]Input[/align]
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.

[align=left]Output[/align]
For each test case, print the value of f(n) on a single line.

[align=left]Sample Input[/align]

1 1 3
1 2 10
0 0 0

[align=left]Sample Output[/align]

2
5

[align=left]Author[/align]
CHEN, Shunbao

[align=left]Source[/align]
ZJCPC2004

1 /*
2     一开始没看到n<=10000000
3     结果疯狂TLE
4
5     最后看了discuss 才知道要找周期
6 */
7 #include<cstdio>
8 #include<iostream>
9 #define MAXN 100000010
10
11 using namespace std;
12
13 int a,b,n;
14
15 int f[1001];
16
17 int main() {
18     while(~scanf("%d %d %d",&a,&b,&n)&&a&&b&&n) {
19         int pos;
20         f[0]=0;f[1]=1;f[2]=1;
21         for(int i=3;i<=100;i++) {
22             f[i]=(f[i-1]*a+f[i-2]*b)%7;
23             if(f[i]==f[2]&&f[i-1]==f[1]) {  //找到周期
24                 pos=i;
25                 break;
26             }
27         }
28 //        printf("%d\n",pos-2);
29         n=n%(pos-2);  //周期为pos-2
30         if(n!=0) printf("%d\n",f
);
31         else printf("%d\n",f[pos-2]);
32     }
33     return 0;
34 }

代码

[align=left]Recommend[/align]
JGShining | We have carefully selected several similar problems for you: 1008 1021 1019 1002 1003