给一个数组,每次操作可以把相邻的两个素数元素进行合并,
合并后的新数为原来的两个数之和,并删除原来两个数。现在希望最终数组的元素数量尽可能少。
输入
第一行 n 代表数组元素个数
第二行 数组的各个元素
4
7 2 2 3
输出
最终的个数
1
package org.example.XC;import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;public class Main {
}class Three {public static void main(String[] args) {Scanner scanner = new Scanner(System.in);int n = scanner.nextInt();List<Integer> numbers = new ArrayList<>();// 读取数组元素for (int i = 0; i < n; i++) {numbers.add(scanner.nextInt());}// 关闭Scannerscanner.close();for (int i = 0; i < numbers.size(); ) {if (isSu(numbers.get(i))) { // if (numbers.get(i)==2) {// ///if (i > 0 && isSu(numbers.get(i - 1)) && isSu(numbers.get(i - 1) + numbers.get(i))) {numbers.set(i - 1, numbers.get(i) + numbers.get(i - 1));numbers.remove(i);} else if (i < numbers.size() - 1 && isSu(numbers.get(i + 1)) && isSu(numbers.get(i) + numbers.get(i + 1))) {numbers.set(i, numbers.get(i) + numbers.get(i + 1));numbers.remove(i + 1);} else {i++;}} else {i++;}}for (int i = 0; i < numbers.size() - 1; i++) {if (isSu(numbers.get(i)) && isSu(numbers.get(i + 1))) {numbers.set(i, numbers.get(i) + numbers.get(i + 1));numbers.remove(i + 1);}}System.out.println(numbers.size());}static boolean isSu(int num) {if (num <= 1) return false;for (int i = 2; i <= Math.sqrt(num); i++) {if (num % i == 0) {return false;}}return true;}
}
第二种
如数组 7 2 2 3,当遇到第一个二时,是不能进行合并的,为了得到最短的结果,我们会使用第二个2和3进行合并,这时,我们就要回退到上一个邻接判断是否为2,再以此个2进行左右合并判断。
第一种方法也可以解决这种问题,但要每个元素进行左右判断,更费时。
class Three2 {public static void main(String[] args) {Scanner scanner = new Scanner(System.in);int n = scanner.nextInt();List<Integer> numbers = new ArrayList<>();// 读取数组元素for (int i = 0; i < n; i++) {numbers.add(scanner.nextInt());}// 关闭Scannerscanner.close();for (int i = 0; i < numbers.size(); ) {if (numbers.get(i) == 2) {if (i > 0 && isSu(numbers.get(i - 1)) && isSu(numbers.get(i - 1) + numbers.get(i))) {numbers.set(i - 1, numbers.get(i) + numbers.get(i - 1));numbers.remove(i);i--;} else if (i < numbers.size() - 1 && isSu(numbers.get(i + 1)) && isSu(numbers.get(i) + numbers.get(i + 1))) {numbers.set(i, numbers.get(i) + numbers.get(i + 1));numbers.remove(i + 1);i--; //回退 看前一位是不是没用使用过的2} else {i++;}} else {i++;}}for (int i = 0; i < numbers.size() - 1; i++) {if (isSu(numbers.get(i)) && isSu(numbers.get(i + 1))) {numbers.set(i, numbers.get(i) + numbers.get(i + 1));numbers.remove(i + 1);}}System.out.println(numbers.size());}static boolean isSu(int num) {if (num <= 1) return false;for (int i = 2; i <= Math.sqrt(num); i++) {if (num % i == 0) {return false;}}return true;}
}