2025 年第十六届蓝桥杯省赛 JavaB 组真题详解

最终答案为 409876661809331。

试题 C：电池分组

核心考点在于位运算中的异或性质。若两组数字的异或和相等，则总异或和必为 0。因此只需遍历数组计算所有元素的异或值，若结果为 0 则输出 YES，否则 NO。

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int t = scan.nextInt();
        while (t-- > 0) {
            int n = scan.nextInt();
            int ans = 0;
            for (int i = 0; i < n; i++) {
                ans ^= scan.nextInt();
            }
            if (ans == 0) {
                System.out.println("YES");
            } else {
                System.out.println("NO");
            }
        }
    }
}

试题 D：魔法科考试

涉及素数判断与集合去重。朴素判断素数的时间复杂度较高，易导致超时。建议采用埃氏筛法预处理素数表，将查询复杂度降至 O(1)。同时使用 Set 结构记录已统计的和，避免重复计数。

import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;

public class Main {
    static int[] prime = new int[20005];
    static boolean[] st = new boolean[20005];
    static int cnt = 0;
    static Set<Integer> set = new HashSet<>();
    static int num;

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        int m = scan.nextInt();
        int[] a = new int[n];
        int[] b = new int[m];
        for (int i = 0; i < n; i++) a[i] = scan.nextInt();
        for (int i = 0; i < m; i++) b[i] = scan.nextInt();

        num = n + m;
        creat();

        int ans = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                int s = a[i] + b[j];
                if (s <= num && set.contains(s)) {
                    ans++;
                    set.remove(s);
                }
            }
        }
        System.out.println(ans);
    }

    public static void creat() {
        for (int i = 2; i <= num; i++) {
            if (!st[i]) {
                prime[cnt++] = i;
                set.add(i);
            }
            for (int j = 0; j < cnt; j++) {
                if (i * prime[j] > num) break;
                st[i * prime[j]] = true;
                if (i % prime[j] == 0) break;
            }
        }
    }
}

试题 E：爆破

几何距离计算问题。两圆最短距离等于圆心距减去两圆半径之和。若结果小于等于 0，说明相交，距离记为 0。需注意坐标轴对齐时的特殊情况及勾股定理的应用。

import java.util.Scanner;

public class Main {
    static int[][] g;

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        g = new int[n + 1][3];
        for (int i = 1; i <= n; i++) {
            g[i][0] = scan.nextInt();
            g[i][1] = scan.nextInt();
            g[i][2] = scan.nextInt();
        }

        double[][] dp = new double[n + 1][n + 1];
        double ans = 0;
        for (int i = 1; i <= n; i++) {
            double min = Double.MAX_VALUE;
            for (int j = 1; j <= n; j++) {
                if (i == j) continue;
                dp[i][j] = func(i, j);
                min = Math.min(min, dp[i][j]);
            }
            ans += min;
        }
        System.out.printf("%.2f", ans);
    }

    public static double func(int i, int j) {
        int lenx = Math.abs(g[i][0] - g[j][0]);
        int leny = Math.abs(g[i][1] - g[j][1]);
        double dist = Math.sqrt(lenx * lenx + leny * leny);
        double result = dist - (g[i][2] + g[j][2]);
        return result <= 0 ? 0 : result;
    }
}

试题 F：数组翻转

本质是寻找两个最大的连续相同区间进行合并。通过遍历统计每个数值的最长连续段长度，取前两大值相乘累加即可得到最大收益。

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        int[] a = new int[n];
        int maxVal = 0;
        for (int i = 0; i < n; i++) {
            a[i] = scan.nextInt();
            maxVal = Math.max(maxVal, a[i]);
        }

        boolean[] visited = new boolean[maxVal + 1];
        long ans = 0;
        for (int i = 0; i < n; i++) {
            if (visited[a[i]]) continue;

            int max1 = 0, max2 = 0, cnt = 0;
            for (int j = i; j < n; j++) {
                if (a[j] == a[i]) {
                    cnt++;
                } else {
                    if (cnt > max1) { max2 = max1; max1 = cnt; }
                    else if (cnt > max2) { max2 = cnt; }
                    cnt = 0;
                }
            }
            if (cnt > max1) { max2 = max1; max1 = cnt; }
            else if (cnt > max2) { max2 = cnt; }

            visited[a[i]] = true;
            ans = Math.max(ans, 1L * a[i] * (max1 + max2));
        }
        System.out.println(ans);
    }
}

试题 G：2 的幂

暂未提供详细解法，建议关注位运算技巧或数学规律。

试题 H：研发资源分配

贪心策略应用。目标是最大化 A 部门收益并最小化 B 部门消耗。优先匹配比当前等级大 1 的资源，若无则选最小可用资源，最后处理相等抵消情况。

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int[] b = new int[n];
        for (int i = 0; i < n; i++) {
            b[i] = scanner.nextInt();
        }

        boolean[] used = new boolean[n + 1];
        int sumA = 0;
        int sumB = 0;

        for (int i = 0; i < n; i++) {
            int index = -1;
            for (int j = 1; j <= n; j++) {
                if (j > b[i] && !used[j]) {
                    index = j;
                    used[j] = true;
                    sumA += i + 1;
                    break;
                }
            }

            if (index == -1) {
                for (int j = 1; j <= n; j++) {
                    if (j < b[i] && !used[j]) {
                        index = j;
                        used[j] = true;
                        sumB += i + 1;
                        break;
                    }
                }
            }

            if (index == -1 && !used[b[i]]) {
                used[b[i]] = true;
            }
        }
        System.out.println(sumA - sumB);
    }
}

总结

本次竞赛侧重逻辑思维与算法优化。建议在考场上先尝试暴力解获取部分分数，再根据数据规模进行剪枝或优化。注意数据类型选择及边界条件处理，避免因细节失误丢分。