HDOJ Number Sequence（java）

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Number Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 98960 Accepted Submission(s): 23770

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 31 2 100 0 0

Sample Output

25
AC代码：利用递推公式去求解肯定会超时，解决本问题的关键之处在于问题是对于7求余，使得所得的解空间一定在0-6范围内，又f[1]，f[2]值给出，找出其循环的位置就可（借鉴了下大牛的代码）
import java.util.Scanner;import java.util.Vector;public class Main {public static void main(String[] args) {Scanner in = new Scanner(System.in);boolean[][] flag = new boolean[7][7];//统计出现首次循环的标志Vector<Integer> v = new Vector<Integer>();while(in.hasNext()){int a = in.nextInt();int b = in.nextInt();int n = in.nextInt();v.clear();if(a==0 && b==0 && n==0)break;for(int i=0; i<7; i++){for(int j=0; j<7; j++){flag[i][j] = false;}}v.add(1);v.add(1);flag[1][1] = true;int count = 1,f;//count记录重复元素之前所包含的元素个数while(true){ f = (a*v.get(count)%7 + b*v.get(count - 1)%7)%7; v.add(f); ++count; if(flag[v.get(count)][v.get(count-1)] == true)//出现相邻重复元素时跳出循环，否则将对应的位置设置为true break; else flag[v.get(count)][v.get(count-1)] = true;}count = count-1;if(n <= count)System.out.println(v.get(n-1));else{int j;for(j=0 ;; j++){if(v.get(j) == v.get(count) && v.get(j+1) == v.get(count+1))//获得首次出现循环元素的下标jbreak;} n = (n - j)%(count - j); if(n == 0) n = count - j; n += j; System.out.println(v.get(n-1));}}}}

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