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如果横竖差值最小的话

两者要么相等，要么差一

令 e1 = n / ((k + 1)/2+1), e2 = n / (k/2 + 1)

则 s = e1 * e2

这样很好的兼顾了k为奇偶的情况

import java.io.*;
import java.util.*;public class Main {public static void main(String[] args) {Scanner sc = new Scanner(new BufferedInputStream(System.in));int n = sc.nextInt();int k = sc.nextInt();double e1 = n * 1.0 / ((k + 1)/2 + 1);double e2 = n * 1.0 / (k/2 + 1);double s = e1 * e2;System.out.printf("%.2f\n", s);}}

这题字符串和操作次数较小，然后可以暴力模拟

如果操作数很多的话，可能需要借助数据结构来维护增量

因为这里面有明显的区间操作.

import java.io.BufferedInputStream;
import java.util.Scanner;public class Main {public static void main(String[] args) {Scanner sc = new Scanner(new BufferedInputStream(System.in));int n = sc.nextInt();int q = sc.nextInt();char[] str = sc.next().toCharArray();for (int i = 0; i < q; i++) {int l = sc.nextInt() - 1, r = sc.nextInt() - 1;int d = r - l + 1;char[] str2 = new char[str.length + d];// 头部System.arraycopy(str, 0, str2, 0,l);// 中间的doublefor (int j = 0; j < d; j++) {str2[l + j * 2] = str2[l + j * 2 + 1] = str[j + l];}// 尾巴System.arraycopy(str, r + 1, str2, r + 1 + d, str.length - (r + 1));str = str2;}System.out.println(new String(str));}}

这题很有意思，先来看一个显而易见的结论

• k == 1, 则结果为 最大值 - 最小值
• k == n, 则结果必然为 0

如果核心的焦点在于， k在两者之间时，如何求解

一开始猜了一个，从收益最大(差值减少梯度)的角度去贪心，结果WA，而且得分不高

一度没辙，后面仔细分析了下，感觉可以枚举最大的没有被选中的项

如果选中某一个项为最大值，那比它小，而离的越近必然被保留，所以k的选择一定分布在前后缀.

所以思路是

• 排序
• 枚举未被选中的最大值
• 利用前后缀优化加速

import java.io.*;
import java.util.*;public class Main {public static void main(String[] args) {Scanner sc = new Scanner(new BufferedInputStream(System.in));int n = sc.nextInt(), k = sc.nextInt();int[] arr = new int[n];for (int i = 0; i < n; i++) {arr[i] = sc.nextInt();}Arrays.sort(arr);if (k == 1) {System.out.println(arr[n - 1] - arr[0]);} else if (k == n) {System.out.println(0);} else {double ans = arr[n - 1] - arr[0];// 前后缀拆解long[] suf = new long[n + 1];for (int j = n - 1; j >= 0; j--) {suf[j] = suf[j + 1] + arr[j];}long pre = 0;// 假设这个元素没被选中for (int i = 0; i < n; i++) {if (i > k) break;long acc = pre + suf[n + i - k];double avg = acc * 1.0 / k;double m1 = Math.max(avg, arr[n + i - k - 1]);double m2 = Math.min(avg, arr[i]);ans = Math.min(ans, m1 - m2);pre += arr[i];}System.out.println(ans);}}}

经典套路题

二分最大重量，然后check逻辑中跑最短路(Dijkstra)进行验证

import java.io.*;
import java.util.*;public class Main {static long inf = Long.MAX_VALUE / 10;static boolean check(int n, List<int[]> []g, int limit, int h) {long[] res = new long[n];Arrays.fill(res, inf);PriorityQueue<long[]> pq = new PriorityQueue<>(Comparator.comparing(x -> x[1]));pq.offer(new long[] {0, 0});res[0] = 0;while (!pq.isEmpty()) {long[] cur = pq.poll();int u = (int)cur[0];if (cur[1] > res[u]) continue;for (int[] e: g[u]) {int v = e[0];if (e[1] >= limit && res[v] > res[u] + e[2]) {res[v] = res[u] + e[2];pq.offer(new long[] {v, res[v]});}}}return res[n - 1] <= h;}public static void main(String[] args) {Scanner sc = new Scanner(new BufferedInputStream(System.in));int n = sc.nextInt(), m = sc.nextInt(), h = sc.nextInt();// 二分List<int[]>[]g = new List[n];Arrays.setAll(g, x -> new ArrayList<>());int mz = 0;for (int i = 0; i < m; i++) {int u = sc.nextInt() - 1, v = sc.nextInt() - 1;int w = sc.nextInt(), d = sc.nextInt();g[u].add(new int[] {v, w, d});g[v].add(new int[] {u, w, d});mz = Math.max(mz, w);}int l = 0, r = mz;while (l <= r) {int mid = l + (r - l) / 2;if (check(n, g, mid, h)) {l = mid + 1;} else {r = mid - 1;}}System.out.println(r);}}