携程今年之机试题也20鸣选择+3编程
网易2017校招编程题,网易2017校编程
小易去购买苹果,有有限栽包装,一种植同等口袋8独,一种同等兜子6个,小易要购买n个苹果(不可知多啊非能够少),输出袋子最少的买进方案,如无法正好买至n个,则输出-1.
public class BuyApple {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.print(“输入一个平头:”);
int n = input.nextInt();
int[] min = buyApple(n);
if(min[0] == n){
System.out.println(-1);
}else{
System.out.println(“最少需要”+min[0]+”袋,其中:8个苹果”+min[1]+”袋、6个苹果”+min[2]+”袋”);
}
}
public static int[] buyApple(int n) {
int min = n;
int[] results = {n,0,0};//总袋数,8只底袋数,6只的袋数
if(n%8 == 0){
results[0] = n/8;
results[1] = n/8;
}else if(n%6 == 0){
if(n/6 < results[0]){
results[0] = n / 6;
results[1] = n / 6;
}
}else{
for(int i = 0; i <= n/8; ++i){
if((n-8*i)%6 == 0){
if((n-8*i)/6 + i < min){
results[0] = (n-8*i)/6 + i;
results[1] = i;
results[2] = (n-8*i)/6;
}
}
}
}
return results;
}
}
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小好去购买苹果,有一定量种植包装,一种植同等口袋8独,一种同等兜子6个,小易要打n个苹果(不能够多吗不克少),输出…
是因为今天最终交给时第三挥毫编程未通过,交卷后想出来的解法这里记录转。
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
//携程3
public class LRU {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int size=Integer.valueOf(sc.nextLine()) ;
List<Integer> list1 = new ArrayList<>();//存key
List<Integer> list2 = new ArrayList<>();//存value
while(sc.hasNext()){
String line =sc.nextLine();
String[] subLine=line.split(" ");
if(subLine[0].equals("p")){
if(list1.contains(Integer.valueOf(subLine[1]))){//如果已经存在,只是将value改变
int k=list1.indexOf(Integer.valueOf(subLine[1]));
list2.set(k,Integer.valueOf(subLine[2]));
}else{//否则放入队列尾
if(list1.size()>=size){
list1.remove(0);
list2.remove(0);
}
list1.add(Integer.valueOf(subLine[1]));
list2.add(Integer.valueOf(subLine[2]));
}
}else{//如果获取值
if(list1.contains(Integer.valueOf(subLine[1]))){
int n =list1.indexOf(Integer.valueOf(subLine[1]));
System.out.println(list2.get(n));
int key=list1.get(n);
int value = list2.get(n);
list2.remove(n);
list1.remove(n);
list1.add(key);
list2.add(value);
}else{
System.out.println(-1);
}
}
}
sc.close();
}
}