【ACM】杭電OJ 1005 Number Sequence (for java)
阿新 • • 發佈:2018-12-01
最開始的程式碼是這樣的,沒說我超時,直接報錯~~
import java.util.Arrays; import java.util.Scanner; public class Main { public static int num(int A,int B,int n) { int[] ans = new int[n + 1]; ans[1] = 1; ans[2] = 1; for(int i = 3; i <= n; i++) { ans[i] = (A * ans[i - 1] + B * ans[i - 2]) % 7; } return ans[n]; } public static void main(String[] args) { // TODO Auto-generated method stub Scanner scanner = new Scanner(System.in); while (scanner.hasNext()) { int A = scanner.nextInt(); int B = scanner.nextInt(); int n = scanner.nextInt(); if (A == 0 && B == 0 && n == 0) { break; } int res = num(A,B,n); System.out.println(res); } } }
後來通過參考別人的程式碼發現是有規律的,最後是對7取餘數,所以,f(n-1),f(n-2)最多各有七種情況(0、1、2、3、4、5、6),所以f(n)最多有7*7=49種情況,所以從一開始到49可能不盡相同,但是從50開始則開始迴圈,f(50)=f(50%49)=f(1)。這是解決超時和記憶體不夠的辦法。
import java.util.Arrays; import java.util.Scanner; public class Main { public static int num(int A,int B,int n) { int[] ans = new int[n + 1]; ans[1] = 1; ans[2] = 1; for(int i = 3; i <= n; i++) { ans[i] = (A * ans[i - 1] + B * ans[i - 2]) % 7; } return ans[n]; } public static void main(String[] args) { // TODO Auto-generated method stub Scanner scanner = new Scanner(System.in); while (scanner.hasNext()) { int A = scanner.nextInt(); int B = scanner.nextInt(); int n = scanner.nextInt(); if (A == 0 && B == 0 && n == 0) { break; } int res = num(A,B,n); System.out.println(res); } } }
參考資料:
https://blog.csdn.net/CSDN___CSDN/article/details/82933844