华泰证券8.28笔试题解代码 为啥第一题我的代码输入的nextLine()报错？？？？ 逻辑是没有问题的，在本地的idea上是可以ac的…… 气死 代码如下，求大佬指正： public static class Main { public static void main(String[] args) { Scanner in = new Scanner(System.in); int k = 1; while (true){ boolean over = false;//是-1？ boolean isZero = false;//是0？ int start = 0; List<List<Integer>> arr = new ArrayList<>(); while (true){ String[] str = in.nextLine().split(" "); if (str[0].equals("") || str.length == 0) { over = true;//空行不管 break; } if (str[0].charAt(0) == '-'){ break;//out } for (int i = 0; i < str.length; i++) { List<Integer> tmp = new ArrayList<>();//一对 int from = str[i].charAt(0) - '0'; if (from == 0) { isZero = true; break; } int to = str[i].charAt(2) - '0'; tmp.add(0); tmp.add(from); tmp.add(to); arr.add(tmp); start = to;//随便选个点 } if (isZero) break;//遇到0，结束 } if (isZero){//树OK？ huaTai23Pro1.Main.Graph graph = generatGrap(arr); boolean ans = bfs(graph.nodes.get(1));//拿起始点 System.out.println(ans ? "Case "+ k +" is a tree" : "Case "+ k +" is not a tree"); } if (over) { k++;//计数增加 continue; } } } //图，BFS给一个图的节点，然后开始遍历，而不是给整张图 public static boolean bfs(huaTai23Pro1.Main.Node head){ if (head == null) return true; Queue<huaTai23Pro1.Main.Node> queue = new LinkedList<>(); HashSet<huaTai23Pro1.Main.Node> set = new HashSet<>(); queue.add(head); set.add(head); boolean ans = true;//默认是树 while (!queue.isEmpty()){ head = queue.poll(); if(head.in > 1) { ans = false; break;//不行了就 } //全部直接邻居 for(huaTai23Pro1.Main.Node next:head.nexts){ if (!set.contains(next)){ //没遍历过才能继续装 queue.add(next); set.add(next); } } } //就这么简单 return ans; } //基础的数据类型，得自个写 //边--和节点是相互渗透的俩数据结构 public static class Edge{ public int weight; public huaTai23Pro1.Main.Node from; public huaTai23Pro1.Main.Node to; public Edge(int w, huaTai23Pro1.Main.Node a, huaTai23Pro1.Main.Node b){ weight = w; from = a; to = b; } } //节点 public static class Node{ public int in; public int out; public int value; public ArrayList<huaTai23Pro1.Main.Node> nexts; public ArrayList<huaTai23Pro1.Main.Edge> edges;//这里是相互渗透的，既然有邻居，就右边 public Node(int v){ value = v;//只需要给这么一个value即可 in = 0; out = 0; nexts = new ArrayList<>(); edges = new ArrayList<>(); } } //图结构玩起来，图右边，节也有边 public static class Graph{ public HashMap<Integer, huaTai23Pro1.Main.Node> nodes;//v,Node public HashSet<huaTai23Pro1.Main.Edge> edges; public Graph(){ nodes = new HashMap<>();//一般节点有value，对应包装袋，都是用哈希表玩的，并查集就是这么玩的 edges = new HashSet<>(); } public int getNodeNum(){ return nodes.size(); } public int getEdgeNum(){ return edges.size(); } } //将非标准图结构转化为左神标准图结构 public static huaTai23Pro1.Main.Graph generatGrap(List<List<Integer>> matrix){ if (matrix == null || matrix.size() == 0) return null; //matrix== //[1,1,2] //[2,2,3] //[3,3,1],w,f,t //挨个遍历行 huaTai23Pro1.Main.Graph graph = new huaTai23Pro1.Main.Graph(); for (int i = 0; i < matrix.size(); i++) { //建节点和边，然后装图，将节点的入度，出度，边和邻居放好 int weight = matrix.get(i).get(0); int from = matrix.get(i).get(1); int to = matrix.get(i).get(2); //图中没节点，建，否则不必重复搞 if (!graph.nodes.containsKey(from)) graph.nodes.put(from, new huaTai23Pro1.Main.Node(from)); if (!graph.nodes.containsKey(to)) graph.nodes.put(to, new huaTai23Pro1.Main.Node(to)); huaTai23Pro1.Main.Node fromNode = graph.nodes.get(from); huaTai23Pro1.Main.Node toNode = graph.nodes.get(to);//有就拿出来 huaTai23Pro1.Main.Edge edge = new huaTai23Pro1.Main.Edge(weight, fromNode, toNode);//建边 graph.edges.add(edge); fromNode.out++; toNode.in++; fromNode.nexts.add(toNode); fromNode.edges.add(edge);//除了入度说终点，其余都是说原点 } return graph; } } 第二题ac class Solution: def upper_bound_(self , n , v , a ): if a[-1] < v: return len(a) + 1 l, r = 0, len(a) - 1 while l < r: mid = l + ((r - l)>>1) if a[mid] < v: l = mid + 1 else: r = mid return l + 1 第三题ac了60% public static class Main { public int a = 10; public static void main(String[] args) { Scanner in = new Scanner(System.in); while (in.hasNextInt()){ int n = in.nextInt(); int[] ability = new int[n]; for (int i = 0; i < n; i++) { ability[i] = in.nextInt(); } int k = in.nextInt(); int d = in.nextInt(); long[][] maxProduct = new long[n][k]; long[][] minProduct = new long[n][k]; for (int i = 0; i < n; i++) { maxProduct[i][0] = ability[i]; minProduct[i][0] = ability[i]; } long max = Long.MIN_VALUE; for (int i = 0; i < n; i++) { for (int j = 1; j < k; j++) { for (int p = i - 1; p >= Math.max(i - d, 0); p--) { maxProduct[i][j] = Math.max(maxProduct[i][j], maxProduct[p][j - 1]); maxProduct[i][j] = Math.max(maxProduct[i][j], maxProduct[p][j - 1] * ability[i]); minProduct[i][j] = Math.min(minProduct[i][j], minProduct[p][j - 1]); minProduct[i][j] = Math.min(minProduct[i][j], minProduct[p][j - 1] * ability[i]); } } max = Math.max(max, maxProduct[i][k - 1]); } System.out.println(max); } } }
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