Codeforces1841D Pairs of Segments

最新推荐文章于 2025-06-06 23:25:37 发布

原创

最新推荐文章于 2025-06-06 23:25:37 发布

· 832 阅读

16 ·

版权

文章标签：

#c++ #算法 #动态规划

中文题面

如果存在至少一个 $x$ ，使得 $l_1 \le x \le r_1$ 和 $l_2 \le x \le r_2$ 相交，则两条线段 $l_1, r_1]$ 和 $l_2, r_2]$ 相交。

如果 $k$ 是偶数，并且可以将这个数组中的元素分成 $\frac{k}{2}$

最低0.47元/天解锁文章

确定要放弃本次机会？

福利倒计时

: :

立减 ¥

普通VIP年卡可用

立即使用

特别萌新的小白

关注关注

12
点赞
踩
16

收藏

觉得还不错? 一键收藏
0
评论
分享

复制链接

分享到 QQ

分享到新浪微博

扫一扫
举报

举报

专栏目录

[codeforces1207F]Remainder Problem

dxyinme-Zz

09-01

305

time limit per test : 4 seconds memory limit per test : 512 megabytes 分数：2100 You are given an array aaa consisting of 500000 integers (numbered from 1 to 500000). Initially all elements of aaa are ze...

CodeForces 289A Polo the Penguin and Segments

平行空间

07-25

911

Polo the Penguin and SegmentsDescriptionLittle penguin Polo adores integer segments, that is, pairs of integers [l; r](l ≤ r).He has a set that consists of n integer segments: [l1; r1], [l2; r2], …, [l

参与评论您还未登录，请先登录后发表或查看评论

Codeforces：All Pairs Segments

2402_86265887的博客

12-02

2234

这是一道1200分的cf题，做了三天，思维题还是不太行，最后还是参考了一个大佬的思路才AC。(我想吐槽一下，这个分数是不是标错了啊，难度远比1200高，呜呜呜～)这道题还是挺有难度的，考的就是模拟，模拟之后还要想该怎样去优化，到最后还是把整个思路换了才AC，将遍历数换成遍历两数之间的区间对我这种蒟蒻还是挺有难度的。，学了一招，以后不管写什么程序都要加上这个，但别忘了将int main()改成signed main()。

D. Pairs of Segments

zzzyyzz_的博客

07-12

125

D. Pairs of Segments

D. Pairs of Segments(最大不相交区间数量)

m0_64158084的博客

06-21

448

给定一个由n个线段组成的数组[[l1,r1],[l2,r2],…如果存在至少一个x满足l1≤x≤r1且l2≤x≤r2，则称线段[l1,r1]和[l2,r2]相交。第一个元素（线段[2,4]）和第三个元素（线段[2,4]）;第二个元素（线段[9,12]）和第五个元素（线段[10,13]）;第四个元素（线段[7,7]）和第六个元素（线段[6,8]）。每个测试用例包含两行，第一行是整数n(2≤n≤2000)，表示数组中线段的个数。不同对之间的线段互不相交。可以看到，每对中的线段相交，不同对之间的线段不相交。

CodeForces - 895B XK Segments

2301_80683422的博客

03-20

236

2）[ a[i] , a[j] ]区间里的数，有k个x的倍数。

Codeforces 193.D Two Segments

weixin_30840253的博客

02-19

101

D. Two Segments time limit per test 5 seconds memory limit per test 256 megabytes input standard input output standard output Nick has some permutation consisting ofpinte...

codeforces 193 D Two Segments

weixin_30800987的博客

01-17

discription Nick has some permutation consisting ofpintegers from1ton. A segment[l, r](l ≤ r) is a set of elementspisatisfyingl ≤ i ≤ r. Nick calls a pair of segments[a0, a1]and[b0, b1]...

C - All Pairs Segments

yingxueyuhanling的博客

12-05

120

【代码】C - All Pairs Segments。

Div3 codeforces1005E1 Median on Segments (Permutations Edition)

Amo丶

07-11

466

E1. Median on Segments (Permutations Edition)time limit per test3 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou are given a permutation p1,p2,…,pnp1,p2,…,pn. A p...

Codeforces 895.B XK Segments

weixin_30501857的博客

12-16

117

B. XK Segments time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output While Vasya finished eating his piece of...

Codeforces Round 975 (Div. 2)(A~F)

weixin_61825750的博客

09-28

2633

疯狂掉分中

CodeForces - 1042D Petya and Array （树状数组+离散化）

柳予欣的三味书屋~

12-02

331

Petya has an array aa consisting of nn integers. He has learned partial sums recently, and now he can calculate the sum of elements on any segment of the array really fast. The segment is a non-empty ...

STL 库基础概念与示例

2302_76213070的博客

06-04

320

自动去重并按字符 ASCII 值排序，插入重复元素无效。按插入顺序存储，需手动排序，支持下标和迭代器遍历。简化值的修改，键唯一（重复插入同一键会更新值）。

C#和C++在编译过程中的文件区分

weixin_50600749的博客

06-04

256

示例： math_utils.cpp --编译--> math_utils.lib。示例： math_utils.cpp --编译--> math_utils.dll。3. 创建类库项目，定义，实现后编译，就生成了.dll文件，如MyLib.dll。2. .cpp是源文件（Source File）2. .cs编译后生成一个.dll文件。把.cpp文件编译成 .lib文件。把.cpp文件编译成 .dll文件。静态库是“编译时拷进程序里的”3.2 动态库 .dll。3.1 静态库 .lib。

C/C++ 面试复习笔记（4）

最新发布

apolloyhl的博客

06-06

881

内存颠簸是指系统在频繁地进行内存页的换入换出操作，导致大量的磁盘 I/O，系统性能急剧下降的现象。当系统的物理内存不足，而进程对内存的需求又不断变化时，就容易出现内存颠簸。例如，一个进程频繁地访问不同的内存页，而这些页在内存中频繁地被换出和换入。在 Linux 系统中，可以通过以下方法避免或解决内存颠簸问题：增加物理内存：如果物理内存不足是导致内存颠簸的主要原因，增加内存条可以直接提高系统的可用内存，减少页交换操作。

极限复习c++

2302_76213070的博客

06-05

1951

快速过一遍必背代码（类、多态、模板）和简答题答案，强化肌肉记忆！祝考试顺利，代码全 AC！

c++ openssl 使用 DES（数据加密标准）进行加密和解密的基本操作

qq_43420088的博客

06-05

906

使用 DES（数据加密标准）进行加密和解密的基本操作，重点展示了 ECB 和 CBC 模式，并且通过篡改密文的方式来进行攻击。下面是对每个部分的详细解析。

加法c++

D_31415926的博客

06-06

351

给定 a 和 b 分别作为被加数和加数，而且限定 a 是 4 位数字，b 是 3 位数字，都是正整数。请你模拟以上加法过程进行输出。如果最后 a+b 的结果是 5 位数，为了保证对齐，需要在被加数和加数前面加空格。只需判断a+b的位数（注：四位数+三位数只可能是四位数或五位数，因为1000+100=1100，9999+999=10998），当a+b为四位数或五位数时输出特定的格式即可。输入两个正整数 a 和 b。

codeforces 1487D

02-19

### Codeforces 1487D Problem Solution The problem described involves determining the maximum amount of a product that can be created from given quantities of ingredients under an idealized production process. For this specific case on Codeforces with problem number 1487D, while direct details about this exact question are not provided here, similar problems often involve resource allocation or limiting reagent type calculations. For instance, when faced with such constraints-based questions where multiple resources contribute to producing one unit of output but at different ratios, finding the bottleneck becomes crucial. In another context related to crafting items using various materials, it was determined that the formula `min(a[0],a[1],a[2]/2,a[3]/7,a[4]/4)` could represent how these limits interact[^1]. However, applying this directly without knowing specifics like what each array element represents in relation to the actual requirements for creating "philosophical stones" as mentioned would require adjustments based upon the precise conditions outlined within 1487D itself. To solve or discuss solutions effectively regarding Codeforces' challenge numbered 1487D: - Carefully read through all aspects presented by the contest organizers. - Identify which ingredient or component acts as the primary constraint towards achieving full capacity utilization. - Implement logic reflecting those relationships accurately; typically involving loops, conditionals, and possibly dynamic programming depending on complexity level required beyond simple minimum value determination across adjusted inputs. ```cpp #include <iostream> #include <vector> using namespace std; int main() { int n; cin >> n; vector<long long> a(n); for(int i=0;i<n;++i){ cin>>a[i]; } // Assuming indices correspond appropriately per problem statement's ratio requirement cout << min({a[0], a[1], a[2]/2LL, a[3]/7LL, a[4]/4LL}) << endl; } ``` --related questions-- 1. How does identifying bottlenecks help optimize algorithms solving constrained optimization problems? 2. What strategies should contestants adopt when translating mathematical formulas into code during competitive coding events? 3. Can you explain why understanding input-output relations is critical before implementing any algorithmic approach? 4. In what ways do prefix-suffix-middle frameworks enhance model training efficiency outside of just tokenization improvements? 5. Why might adjusting sample proportions specifically benefit models designed for tasks requiring both strong linguistic comprehension alongside logical reasoning skills?

Codeforces1841D Pairs of Segments

tags

中文题面