转载声明:原文转自http://www.cnblogs.com/xiezie/p/5502918.html
JAVA语言实现:
拿到题目第一反应是简单地实现递归:
import java.util.*; import java.io.*; public class Main{ public static void main(String[] arg){
Main m = new Main();
Scanner scan = new Scanner(new BufferedInputStream(System.in));
while(scan.hasNextInt()){
int a = scan.nextInt();
int b = scan.nextInt();
int n = scan.nextInt();
if(a==0&&b==0&&n==0){
break;
}
int result = m.f(a,b,n);
System.out.println(result);
}
scan.close();
} public int f(int a, int b ,int n){
int result = 0;
if(n>2){
result = (a*f(a,b,n-1) + b*f(a,b,n-2))%7;
}else{
result = 1;
}
return result;
}
}
在实现时 还是考虑到JVM能够递归的深度,所以报出Memory Limit Exceeded的错误 让我觉得很正常,而且实现上不优雅~
第二次是将递归改为循环 自我感觉良好 结果 报了Time Limit Exceeded的错误
一下让我措手不及~
import java.util.*; import java.io.*; public class Main{ public static void main(String[] arg){
Scanner scan = new Scanner(new BufferedInputStream(System.in));
while(scan.hasNextInt()){
int a = scan.nextInt();
int b = scan.nextInt();
int n = scan.nextInt();
if(a==0&&b==0&&n==0){
break;
}
int result = 1;
int result1 = 1,result2;
if(n>2){
int m = n+1;
for(int i = 3 ; i != m ; i ++ ){
result2 = result1;
result1 = result;
result = (a*result1 + b * result2 )%7;
}
}
System.out.println(result);
}
scan.close();
}
最后还是得分析算法 - -
这个f(n)得出的数字 都在0-6,网上的说法是 A%7,B%7 所以49个数字一个周期
这个有考虑过 但是实现时 还是没去考虑这个 - - 还是太懒~
于是 代码实现上是
import java.util.*; import java.io.*; public class Main{ public static void main(String[] arg){
Scanner scan = new Scanner(new BufferedInputStream(System.in));
while(scan.hasNextInt()){
int a = scan.nextInt();
int b = scan.nextInt();
int n = scan.nextInt();
if(a==0&&b==0&&n==0){
break;
}
int result = 1;
int result1 = 1,result2;
int i = 3;
int m = n%49 + 1;
if(m!=1&&m!=2){ while(i++<m){
result2 = result1;
result1 = result;
result = (a*result1 + b * result2 )%7;
}
}
System.out.println(result);
}
scan.close();
}
}