HelloACM_ICPC的博客

蓝桥杯第七届决赛真题JAVA大学C组

HelloACM_ICPC — Wed, 26 May 2021 17:26:14 +0800

蓝桥杯第七届决赛真题JAVA大学C组 #include #include using namespace std; bool vis[1005]; int main() { int ans = 0; for (int i = 0; i <= 1000; i ++) { if (!vis[(i * i) % 100]) ans ++; vis[(i * i) % 100] = true; } cout <<

蓝桥杯2016年第七届C++B组

HelloACM_ICPC — Mon, 24 May 2021 18:48:09 +0800

蓝桥杯2016年第七届C++B组 #include #include using namespace std; int main() { int ans = 600; for (int i = 0; i <= 300; i ++) { for (int j = 0; j <= 300; j ++) { if (i * 97 - j * 127 == 1) { ans = min(ans, i +

第十届蓝桥杯大赛软件类决赛Java大学B组

HelloACM_ICPC — Fri, 30 Apr 2021 16:49:09 +0800

第十届蓝桥杯大赛软件类决赛Java大学B组 #include #include #include #include #include using namespace std; int n = 30, m = 50; char mp[55][55]; int cnt = 0; //string str = "LANNQIAO"; string

ACM训练赛第19场补题

HelloACM_ICPC — Thu, 22 Apr 2021 20:30:15 +0800

ACM训练赛第19场补题问题 A: At Least Average 时间限制: 1 Sec 内存限制: 128 MB 题目描述 Blake plays a video game that has n levels. On each level, she can earn in between 0 and 10 points, inclusive. The number of points she gets on a level must be an integer. She has set up a

ACM训练赛第17场

HelloACM_ICPC — Wed, 14 Apr 2021 16:32:47 +0800

ACM训练赛第17场问题 A: Bing Bong to Bong Bong 时间限制: 2 Sec 内存限制: 128 MB 题目描述 Bing Bong, the pink imaginary friend from Riley’s childhood, is lonely.He wants to preserve his memory, but he wants to try to hide his name in the stories. He wants you to help him do

ACM训练赛第16场

HelloACM_ICPC — Mon, 12 Apr 2021 17:36:58 +0800

ACM训练赛第16场问题 F: Exam Manipulation 时间限制: 1 Sec 内存限制: 128 MB 题目描述 A group of students is taking a True/False exam. Each question is worth one point. You, as their teacher, want to make your students look as good as possible—so you cheat! (I know, you would

Linux环境下输出Hello World

HelloACM_ICPC — Fri, 02 Apr 2021 17:05:05 +0800

Linux环境下输出Hello World 进入home目录 cd /home 创建C++目录 mkdir C++ 进入C++目录下 cd C++ 创建C语言文件 touch HelloWorld.c 编辑C语言文件 vim HelloWorld.c 进入后按键盘上的i进入insert模式敲入C语言代码按住Esc键退出写入模式保存退出命令 :wq 生成可执行文件 gcc HelloWorld.c -o HelloWorld 运行HelloWorld文件 ./HelloWorld

贪心

HelloACM_ICPC — Thu, 01 Apr 2021 20:44:13 +0800

贪心 import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); int[] a = new int[n + 5]; for (int i = 0; i < n; i ++) { a[i] = in.nextInt(); } int

双指针与BFS

HelloACM_ICPC — Wed, 31 Mar 2021 19:25:28 +0800

双指针与BFS import java.util.ArrayList; import java.util.Collections; import java.util.Scanner; import java.util.Comparator; public class Main { static int[] cnt = new int[100000 + 5]; static boolean[] st = new boolean[100000 + 5]; public static void

枚举、模拟与排序

HelloACM_ICPC — Fri, 26 Mar 2021 22:19:03 +0800

枚举、模拟与排序 import java.util.Scanner; public class Main { static int INF = 0x3f3f3f3f; public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); int[] a = new int[n + 5]; for (int i = 0; i < n; i ++

数学与简单DP

HelloACM_ICPC — Wed, 24 Mar 2021 20:56:58 +0800

数学与简单DP import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); int m = in.nextInt(); System.out.println(n * m - n - m); } } import java.util.Scanner

二分与前缀和

HelloACM_ICPC — Tue, 23 Mar 2021 19:25:03 +0800

二分与前缀和 import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); int q = in.nextInt(); int[] a = new int[n + 5]; for (int i = 0; i < n; i ++) { a[i

递归与递推

HelloACM_ICPC — Mon, 22 Mar 2021 16:40:02 +0800

递归与递推 import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); for (int i = 0; i < 1 << n; i ++) { for (int j = 0; j < n; j ++) { if ((i

蓝桥杯九宫重排

HelloACM_ICPC — Sat, 07 Nov 2020 17:14:58 +0800

蓝桥杯九宫重排 C++ #include #include #include #include #include #include using namespace std; int dx[] = {1, -1, 0, 0}; int dy[] = {0, 0, 1, -1}; int factor[] = {

区间dp模板石子合并

HelloACM_ICPC — Sun, 01 Nov 2020 15:22:26 +0800

石子合并设有N堆石子排成一排，其编号为1，2，3，…，N。每堆石子有一定的质量，可以用一个整数来描述，现在要将这N堆石子合并成为一堆。每次只能合并相邻的两堆，合并的代价为这两堆石子的质量之和，合并后与这两堆石子相邻的石子将和新堆相邻，合并时由于选择的顺序不同，合并的总代价也不相同。例如有4堆石子分别为 1 3 5 2，我们可以先合并1、2堆，代价为4，得到4 5 2，又合并 1，2堆，代价为9，得到9 2 ，再合并得到11，总代价为4+9+11=24；如果第二步是先合并2，3堆，则代价为7，得

01背包优先队列优化

HelloACM_ICPC — Sat, 31 Oct 2020 19:17:11 +0800

01背包优先队列优化设重量为w，权值为v 则单位重量价值为v / w 设有物品1，物品2 若v1 / w1 < v2 / w2 则说明v2的单位重量价值高即v1 * w2 < v2 * w1 于是可重载结构体，使得v2 * w1大的优先级高即： struct Node { LL w, v; Node() {}; Node(LL _w, LL _v) {w = _w, v = _v;} friend bool operator < (Node a, Node b) {

Java函数判定大素数

HelloACM_ICPC — Wed, 28 Oct 2020 17:30:05 +0800

Java函数判定大素数 import java.math.BigInteger; import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner in = new Scanner(System.in); BigInteger x; while(in.hasNextBigInteger()) { x = in.nextBigInteger(); if(x

SPFA最短路Java模板

HelloACM_ICPC — Thu, 15 Oct 2020 19:01:06 +0800

SPFA最短路Java模板 import java.util.LinkedList; import java.util.Queue; public class Main { public static void main(String[] args) { } static int MAXN = 10000 + 10; static int[] head = new int[MAXN]; static int[] nxt = new int[MAXN << 1]; st

ST表

HelloACM_ICPC — Tue, 13 Oct 2020 10:27:02 +0800

ST表模板 ST表求区间最大值模板 import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner in = new Scanner(System.in); int T = in.nextInt(); while(T -- > 0) { int n = in.nextInt(); num = new int[100000 + 10][20]

二进制枚举

HelloACM_ICPC — Fri, 09 Oct 2020 13:07:02 +0800

二进制枚举 ACM中二进制枚举C++1秒大约能枚举22位， for循环不能跑太多，加上子层循环和判断，所以一般范围大概就是在22左右。 int i = 0; i < 1 << 22 1 << 30是1e9 超过int类型需要转换为long long类型 1ll << 30 1ll << 60 二进制枚举范围为0到2^n - 1。例题 import java.util.Scanner; public class Main { // 15个