【藍橋杯】第十六屆藍橋杯 JAVA B組記錄

試題 A: 逃離高塔

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很簡單，簽到題，但是需要注意精度，用int會有溢出風險

答案：202

package lanqiao.t1;import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.io.StreamTokenizer;public class Main {static class io {static InputStreamReader ins = new InputStreamReader(System.in);static StreamTokenizer in = new StreamTokenizer(ins);static BufferedReader br = new BufferedReader(ins);static PrintWriter out = new PrintWriter(System.out);static int readInt() throws IOException {in.nextToken();return (int) in.nval;}}public static void main(String[] args) {long n = 2025;long ans = 0;for (long i = 1; i <= n; i++) {long num = i * i * i;if (num % 10 == 3) {ans++;}}io.out.println(ans);io.out.flush();}
}

試題 B: 消失的藍寶

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讀完一遍題，發現就是求N的最小，這里令兩個日期分別為 a，b，那么就是
（n + a）%b == 0
(n+b)%a==0
第一眼想的是中國剩余定理，但是后面一看可以暴力
可以轉換為求（n+a）%b == 0 ,那么（n+a）一定是 b的倍數，那么就可以轉為為：
求 (kb - 9999) % a == 0 ，把k求出來直接反推就行了]

答案: 409876661809331


package lanqiao.t2;import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.io.StreamTokenizer;public class Main {static class io {static InputStreamReader ins = new InputStreamReader(System.in);static StreamTokenizer in = new StreamTokenizer(ins);static BufferedReader br = new BufferedReader(ins);static PrintWriter out = new PrintWriter(System.out);static int readInt() throws IOException {in.nextToken();return (int) in.nval;}}public static void main(String[] args) {long x = 20250412, y = 20240413;boolean flag = true;for (int i = 9999; i <= 99999999 && flag; i++) {long num = i * y;io.out.println(num);io.out.flush();if ((num - 9999) % x == 0) {io.out.println(i);io.out.flush();break;}}io.out.flush();}}

試題 C: 電池分組

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俺位異或，就是將兩個數轉成二進制，如果相同的位數都為1，那么結果這一位就是1，其余就是0
有兩種想法，一種是統計二進制每一位有多少個，如果最后所有位數都是偶數，那么就是可以的，否則就不行
另外一種就是第一種的簡化，如果兩組的異或結果相同，那么兩個再異或一定為0，所以只需要看是否異或起來結果為0就行


public class Main {static class io {static InputStreamReader ins = new InputStreamReader(System.in);static StreamTokenizer in = new StreamTokenizer(ins);static BufferedReader br = new BufferedReader(ins);static PrintWriter out = new PrintWriter(System.out);public static int readInt() throws IOException {in.nextToken();return (int) in.nval;}}public static void main(String[] args) throws IOException {int t = io.readInt();while (t-- > 0) {int n = io.readInt();int num = 0;for (int i = 1; i <= n; i++) {num = num ^ io.readInt();}if (num == 0) {io.out.println("YES");} else {io.out.println("NO");}io.out.flush();}}
}

試題 D: 魔法科考試

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題意就是用 a數組和b數組進行兩兩組合，如果為質數，并且小于等于n+m，那么就是有效的，問一共有多少種，其實真正的復雜度在于如何去判斷質數，如果暴力那肯定不行，所以需要預處理，用質數篩預處理就行了


public class Main {static class io {static InputStreamReader ins = new InputStreamReader(System.in);static StreamTokenizer in = new StreamTokenizer(ins);static BufferedReader br = new BufferedReader(ins);static PrintWriter out = new PrintWriter(System.out);public static int readInt() throws IOException {in.nextToken();return (int) in.nval;}}static Set<Integer> set = new TreeSet<Integer>();static {int num = 400000;int cnt = 0;int[] st = new int[num + 1];int[] prime = new int[num + 1];for (int i = 2; i <= num; i++) {if (st[i] == 0) {st[i] = 1;prime[cnt++] = i;set.add(i);}for (int j = 0; j < cnt && i * prime[j] <= num; j++) {st[i * prime[j]] = 1;if (i % prime[j] == 0) {break;}}}}public static void main(String[] args) throws IOException {int n = io.readInt(), m = io.readInt();int[] a = new int[n], b = new int[m];for (int i = 0; i < n; i++) {a[i] = io.readInt();}for (int i = 0; i < m; i++) {b[i] = io.readInt();}Arrays.sort(a);Arrays.sort(b);Set<Integer> ans = new HashSet<Integer>();for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {int num = a[i] + b[j];if (num > n + m)break;if (set.contains(num)) {ans.add(num);}}}io.out.print(ans.size());io.out.flush();}
}

試題 E: 爆破

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題意分析：將所有圓鏈接起來，所用的長度最小，首先想到的就是最小生成樹，那么直接用prim或者kruskal算法，那么莫如何轉換為算法的模型呢，如果兩個圓相交，那么這兩個圓心連城的邊就是0，如果未相交，那么就是圓心距離-r1-r2,然后建圖即可


public class Main {static class io {static InputStreamReader ins = new InputStreamReader(System.in);static StreamTokenizer in = new StreamTokenizer(ins);static BufferedReader br = new BufferedReader(ins);static PrintWriter out = new PrintWriter(System.out);public static int readInt() throws IOException {in.nextToken();return (int) in.nval;}}static class Point {int no;int x;int y;int r;public Point(int no, int x, int y, int r) {this.no = no;this.x = x;this.y = y;this.r = r;}}static int N = 5010;static class Edge implements Comparable<Edge> {int a;int b;double c;public Edge(int a, int b, double c) {this.a = a;this.b = b;this.c = c;}@Overridepublic int compareTo(Edge o) {return Double.compare(this.c, o.c);}}static int[] fa;static int find(int x) {return fa[x] == x ? fa[x] : (fa[x] = find(fa[x]));}public static void main(String[] args) throws IOException {int n = io.readInt();fa = new int[n + 1];for (int i = 1; i <= n; i++) {fa[i] = i;}Point[] points = new Point[n + 1];for (int i = 1; i <= n; i++) {points[i] = new Point(i, io.readInt(), io.readInt(), io.readInt());}int cnt = 0;Edge[] edges = new Edge[n * n];for (int i = 1; i <= n; i++) {for (int j = i + 1; j <= n; j++) {Point point1 = points[i], point2 = points[j];int r1 = point1.r, r2 = point2.r;double dis = Math.sqrt((point1.x - point2.x) * (point1.x - point2.x) + (point1.y - point2.y) * (point1.y - point2.y));if (r1 + r2 < dis) {double d = dis - r1 - r2;edges[cnt++] = new Edge(i, j, d);} else {edges[cnt++] = new Edge(i, j, 0);}}}Arrays.sort(edges, 0, cnt);double ans = 0;for (int i = 0; i < cnt; i++) {Edge edge = edges[i];int f1 = find(edge.a), f2 = find(edge.b);if (f1 == f2)continue;ans += edge.c;fa[f1] = f2;}io.out.printf("%.2f", ans);io.out.flush();}
}

試題 F: 數組翻轉

不會，寫了個暴力，應該能騙點分

試題 G: 2 的冪

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寫了個樸素算法，騙分

試題 H: 研發資源分配

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直接模擬

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