abc287d & 1124-1250-1297 & 第332场周赛T1~T3 (6)

Author: gdut-yy

atcoder-beginner/src/main/java/c279/Abc279_d.java

@@ -4,33 +4,32 @@
 import java.util.Scanner;
 
 public class Abc279_d {
+    static long a, b;
+
     public static void main(String[] args) {
         Scanner scanner = new Scanner(System.in, StandardCharsets.UTF_8);
-        long A = scanner.nextLong();
-        long B = scanner.nextLong();
-
-        long left = 0;
-        long right = Math.max(A, B);
-        double lans = f(A, B, right);
-        double rans = f(A, B, left);
-        while (left < right) {
-            long lmid = left + (right - left) / 3;
-            long rmid = right - (right - left) / 3;
-            lans = f(A, B, lmid);
-            rans = f(A, B, rmid);
+        a = scanner.nextLong();
+        b = scanner.nextLong();
 
-            if (lans <= rans) {
-                right = rmid - 1;
+        long l = 0, r = a / b;
+        while (r - l > 2) {
+            long m1 = (l * 2 + r) / 3;
+            long m2 = (l + r * 2) / 3;
+            if (f(m1) > f(m2)) {
+                l = m1;
             } else {
-                left = lmid + 1;
+                r = m2;
             }
         }
-        double res = Math.min(lans, rans);
+        double res = a;
+        for (long i = l; i <= r; i++) {
+            res = Math.min(res, f(i));
+        }
         System.out.printf("%.10f%n", res);
     }
 
-    private static double f(long A, long B, long k) {
-        return (double) B * k + A / Math.sqrt(k + 1);
+    private static double f(long n) {
+        return (double) b * n + (double) a / Math.sqrt(n + 1);
     }
 }
 /*

atcoder-beginner/src/main/java/c287/Abc287_d.java

@@ -0,0 +1,50 @@
+package c287;
+
+import java.nio.charset.StandardCharsets;
+import java.util.Scanner;
+
+public class Abc287_d {
+    public static void main(String[] args) {
+        Scanner scanner = new Scanner(System.in, StandardCharsets.UTF_8);
+        String s = scanner.next();
+        String t = scanner.next();
+        System.out.println(solve(s, t));
+    }
+
+    private static String solve(String s, String t) {
+        int n = s.length();
+        int m = t.length();
+
+        // pre[i] 表示前缀长度为 i 时，前缀是否 match
+        boolean[] pre = new boolean[m + 1];
+        pre[0] = true;
+        for (int i = 1; i <= m; i++) {
+            char ch1 = s.charAt(i - 1);
+            char ch2 = t.charAt(i - 1);
+            pre[i] = pre[i - 1] && (ch1 == ch2 || ch1 == '?' || ch2 == '?');
+            if (!pre[i]) break;
+        }
+
+        // suf[i] 表示后缀长度为 i 时，后缀是否 match
+        boolean[] suf = new boolean[m + 1];
+        suf[0] = true;
+        for (int i = 1; i <= m; i++) {
+            char ch1 = s.charAt(n - i);
+            char ch2 = t.charAt(m - i);
+            suf[i] = suf[i - 1] && (ch1 == ch2 || ch1 == '?' || ch2 == '?');
+            if (!suf[i]) break;
+        }
+
+        String[] res = new String[m + 1];
+        for (int i = 0; i < m + 1; i++) {
+            res[i] = pre[i] && suf[m - i] ? "Yes" : "No";
+        }
+        return String.join(System.lineSeparator(), res);
+    }
+}
+/*
+D - Match or Not
+https://atcoder.jp/contests/abc287/tasks/abc287_d
+
+预处理 前后缀
+ */

atcoder-beginner/src/test/java/c287/Abc287dTests.java

@@ -0,0 +1,33 @@
+package c287;
+
+import base.AbstractOjTests;
+import org.junit.jupiter.api.Test;
+
+import java.io.IOException;
+
+public class Abc287dTests extends AbstractOjTests {
+    public Abc287dTests() {
+        super("/c287/D/");
+    }
+
+    @Test
+    public void example1() throws IOException {
+        super.doSetSystemInOut(INPUT1);
+        Abc287_d.main(null);
+        super.doAssertion(OUTPUT1);
+    }
+
+    @Test
+    public void example2() throws IOException {
+        super.doSetSystemInOut(INPUT2);
+        Abc287_d.main(null);
+        super.doAssertion(OUTPUT2);
+    }
+
+    @Test
+    public void example3() throws IOException {
+        super.doSetSystemInOut(INPUT3);
+        Abc287_d.main(null);
+        super.doAssertion(OUTPUT3);
+    }
+}

atcoder-beginner/src/test/resources/c287/D/input1.txt

@@ -0,0 +1,2 @@
+a?c
+b?

atcoder-beginner/src/test/resources/c287/D/input2.txt

@@ -0,0 +1,2 @@
+atcoder
+?????

atcoder-beginner/src/test/resources/c287/D/input3.txt

@@ -0,0 +1,2 @@
+beginner
+contest

atcoder-beginner/src/test/resources/c287/D/output1.txt

@@ -0,0 +1,3 @@
+Yes
+No
+No

atcoder-beginner/src/test/resources/c287/D/output2.txt

@@ -0,0 +1,6 @@
+Yes
+Yes
+Yes
+Yes
+Yes
+Yes

atcoder-beginner/src/test/resources/c287/D/output3.txt

@@ -0,0 +1,8 @@
+No
+No
+No
+No
+No
+No
+No
+No

leetcode-12/src/main/java/Solution1124.java

@@ -0,0 +1,45 @@
+import java.util.HashMap;
+import java.util.Map;
+
+public class Solution1124 {
+    public int longestWPI(int[] hours) {
+        int n = hours.length;
+        Map<Integer, Integer> sumMap = new HashMap<>();
+        int sum = 0;
+        int max = 0;
+        for (int j = 0; j < n; j++) {
+            // 求解区间分数和大于 0 的最长区间长度
+            sum += hours[j] > 8 ? 1 : -1;
+            if (sum > 0) {
+                // [0, j]
+                max = Math.max(max, j + 1);
+            } else {
+                if (sumMap.containsKey(sum - 1)) {
+                    int i = sumMap.get(sum - 1);
+                    // [0,j] - [0,i] = (i,j]
+                    max = Math.max(max, j - i);
+                }
+            }
+            if (!sumMap.containsKey(sum)) {
+                sumMap.put(sum, j);
+            }
+        }
+        return max;
+    }
+}
+/*
+1124. 表现良好的最长时间段
+https://leetcode.cn/problems/longest-well-performing-interval/
+
+给你一份工作时间表 hours，上面记录着某一位员工每天的工作小时数。
+我们认为当员工一天中的工作小时数大于 8 小时的时候，那么这一天就是「劳累的一天」。
+所谓「表现良好的时间段」，意味在这段时间内，「劳累的天数」是严格 大于「不劳累的天数」。
+请你返回「表现良好时间段」的最大长度。
+提示：
+1 <= hours.length <= 10^4
+0 <= hours[i] <= 16
+
+前缀和 + 哈希
+相似题目: 题目-02. 销售出色区间
+https://leetcode.cn/contest/hhrc2022/problems/0Wx4Pc/
+ */