题型: 签到

节日快乐，出题人也说出来自己的心愿, 祝大家AK快乐!

import java.util.Scanner;public class Main {public static void main(String[] args) {System.out.println("Today AK!");}
}

思路: 模拟

按题意模拟即可，环状结构

import java.io.BufferedInputStream;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
import java.util.stream.Collectors;public class Main {public static void main(String[] args) {Scanner sc = new Scanner(new BufferedInputStream(System.in));int n = sc.nextInt();int[] arr = new int[n];for (int i = 0; i < n; i++) {arr[i] = sc.nextInt();}List<Integer> res = new ArrayList<>();for (int i = 0; i < n; i++) {res.add(arr[(i - 1 + n) % n] + arr[(i + 1) % n]);}System.out.println(res.stream().map(String::valueOf).collect(Collectors.joining(" ")));}}

思路: 思维题

可以贪心逆序从右到左翻转，每次翻转保证最右的非负数多一项，题意保证最多有n次。

所以，必然存在操作序列，使得数组元素全变非负形态。

$$\sum _{i=0}^{i=n-1}abs(a_i)$$

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();long[] arr = new long[n];long sum = 0;for (int i = 0; i < n; i++) {arr[i] = sc.nextLong();sum += Math.abs(arr[i]);}System.out.println(sum);}}

思路: 解方程

假设第一个元素为y, 那么

$$(y+2{)}^2-y^2=4(y+1)=x$$

那么该y为

$$y=x/4-1$$

这边需要保证x是4的倍数，同时y必须为>=1的奇数

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 t = sc.nextInt();while (t-- > 0) {long x = sc.nextLong();if (x % 4 != 0) {System.out.println("No");} else {long y = x / 4 - 1;if (y <= 0 || y % 2 == 0) {System.out.println("No");} else {System.out.println("Yes");System.out.println((y) + " " + (y + 2));}}}}}

思路: 贪心+双指针

感觉这题是这场周赛最难的，但是和历史比，算中等偏下的题

因为三角形中，边的关系需要满足,

$$任意两边之和必须大于第三边$$

这题也是最大二分匹配模型

不过这题有个取巧的地方，就是

从底边出发

如果 $b_i<b_j$, 则选用$b_i$ 比 $b_j$ 的结果不会变差

因此可以对腰边和底边进行排序

然后使用双指针，进行贪心配对

import java.io.BufferedInputStream;
import java.util.Arrays;
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[] arr = new int[n];int[] brr = new int[n];for (int i = 0; i < n; i++) {arr[i] = sc.nextInt();}for (int i = 0; i < n; i++) {brr[i] = sc.nextInt();}Arrays.sort(arr);Arrays.sort(brr);// 双指针int res = 0;int j = 0;for (int i = 0; i < n; i++) {while (j < n && arr[j] * 2 <= brr[i]) {j++;}if (j < n) {j++;res++;}}System.out.println(res);}}

思路: 二分

因为其函数是单调的，呈现单调性

因此用二分是最优解

import java.io.BufferedInputStream;
import java.util.Scanner;public class Main {static boolean check(int x, int k, int m) {double r1 = Math.sqrt(1.0 * x);double r2 = (int)(Math.log(x) / Math.log(k));return r1 + r2 > m;}public static void main(String[] args) {Scanner sc = new Scanner(new BufferedInputStream(System.in));int t = sc.nextInt();while (t-- > 0) {int k = sc.nextInt(), m = sc.nextInt();int l = 1, r = m * m;while (l <= r) {int mid = l + (r - l) / 2;if (check(mid, k, m)) {r = mid - 1;} else {l = mid + 1;}}System.out.println(l);}}}