华为OD机考-素数伴侣-逻辑分析（JAVA 2025B卷）

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;public class PrimeCouple {public static void main(String[] args) throws IOException {/*** 奇数偶数 分组* 循环奇数组 找到偶数组中能够奇数匹配的素数伴侣* 已有伴侣的偶数则让给后来可以重新找到伴侣的*/BufferedReader br = new BufferedReader(new InputStreamReader(System.in));int N = Integer.valueOf(br.readLine());String [] lines = br.readLine().split("\\s");List<Integer> odds = new ArrayList<>(8);List<Integer> evens = new ArrayList<>(8);for(String i : lines){int item = Integer.valueOf(i);if(item % 2 == 1){odds.add(item);}if(item % 2 == 0){evens.add(item);}}int count = 0;int [] matchEvens = new int[evens.size()];for(int odd : odds){boolean [] v = new boolean[evens.size()];if(find(odd,matchEvens,evens,v)){count++;}}System.out.println(count);}static boolean find(int odd,int[] matchEvens,List<Integer> evens,boolean[] v){//遍历偶数//去检查当前传入的奇数能否与偶数匹配for (int i = 0; i < evens.size(); i++){boolean prime = isPrime(odd + evens.get(i));if(prime && !v[i]){v[i] = true;//如果第i个偶数没有伴侣//或者第i个偶数原来有伴侣，并且该伴侣能够重新找到伴侣的话(这里有递归调用)//则奇数x可以设置为第i个偶数的伴侣//这里采用了匈牙利算法，能让则让if(matchEvens[i] == 0 || find(matchEvens[i],matchEvens,evens,v)){matchEvens[i] = odd;return true;}}}return false;}private static boolean isPrime(int n) {if(n == 1) return false;for (int i = 2;i <= (int)Math.sqrt(n); i++){if(n % i == 0)return false;}return true;}
}