Xidian ACM Online Judge. Http:/ acm.xidian.edu.cn/download/seat.xlsx. Http:/ acm.xidian.edu.cn/download/rule.doc. Http:/ acm.xidian.edu.cn/download/ranklist.xlsx. 负责人 12-电院-山水 贪心只能过样例 CF老是FST. Http:/ acm.xidian.edu.cn/download/editorial.pdf. Anything about the Problems, Please Contact Admin: admin.
http://acm.xidian.edu.cn/
TODAY'S RATING
>1,000,000
Date Range
HIGHEST TRAFFIC ON
Friday
LOAD TIME
3.8 seconds
PAGES IN
THIS WEBSITE
6
SSL
EXTERNAL LINKS
21
SITE IP
202.117.120.247
LOAD TIME
3.838 sec
SCORE
6.2
Welcome To Online Judge | acm.xidian.edu.cn Reviews
https://acm.xidian.edu.cn
Xidian ACM Online Judge. Http:/ acm.xidian.edu.cn/download/seat.xlsx. Http:/ acm.xidian.edu.cn/download/rule.doc. Http:/ acm.xidian.edu.cn/download/ranklist.xlsx. 负责人 12-电院-山水 贪心只能过样例 CF老是FST. Http:/ acm.xidian.edu.cn/download/editorial.pdf. Anything about the Problems, Please Contact Admin: admin.
acm.xidian.edu.cn
排名
http://acm.xidian.edu.cn/ranklist.php
Xidian ACM Online Judge. Anything about the Problems, Please Contact Admin: admin.
Welcome To Online Judge
http://acm.xidian.edu.cn/faqs.cn.php
Xidian ACM Online Judge. Xidian ACM Online Judge Online Judge FAQ. Linux 使用 GNU GCC/G. 作为C/C 编译器, Free Pascal. 作为pascal 编译器 ，用 sun-java-jdk1.7. 编译 Java. 对应的编译选项如下:. Gcc Main.c -o Main -fno-asm -O2 -Wall -lm - static -std=c99 -DONLINE JUDGE. G Main.cc -o Main -fno-asm -O2 -Wall -lm - static -DONLINE JUDGE. Fpc Main.pas -oMain -O1 -Co -Cr -Ct -Ci. Javac -J-Xms32m -J-Xmx256m Main.java. Java has 2 more seconds and 512M more memory when running and judging. Free Pascal Compiler version 2.6.2-8 for i386.
Recent Contests from Naikai-contest-spider
http://acm.xidian.edu.cn/recent-contest.php
Xidian ACM Online Judge. BNU Training 2016.08.19 Div. 1. BNU Training 2016.08.19 Div. 2. Codeforces Round #368 (Div. 2). Educational Codeforces Round 16. How Much Can a Nation Save MM. AIMTech Round 3 (Div. 1). AIMTech Round 3 (Div. 2). TCO16 Algorithm Round 3B. DNA Sequencing 2 MM. Ad Infinitum 16 (First timers). TENTATIVE Fishing for Fishermen. TCO16 Online Wildcard Round. Week of Code 23. DataSource:http:/ contests.acmicpc.info/contests.json Spider Author: doraemonok.
Problem Set
http://acm.xidian.edu.cn/problemset.php
Xidian ACM Online Judge. 亮亮做加法 (a.k.a another A B Problem). A Simple Math Problem. A simple math problem 2. A B of lw. Anything about the Problems, Please Contact Admin: admin.
状态
http://acm.xidian.edu.cn/status.php
Xidian ACM Online Judge. Anything about the Problems, Please Contact Admin: admin.
TOTAL PAGES IN THIS WEBSITE
6
MCS - 数一新天地
http://mathlover.info/mcs/mathematics-for-computer-science
前几天谷歌和麻省理工大学发布了2017年版的 mathematics for computer science. 20Deviation from the Mean / 看起来像是各种定理. 2014-2015 ACM-ICPC, NEERC, Moscow Subregional Contest解题报告. Hdu5119 Happy Matt Friends. 发表在 2014-2015 ACM-ICPC, NEERC, Moscow Subregional Contest解题报告. 发表在 BC#12 && CF 375有感。 2014 Multi-University Training Contest 10.
数一新天地 | Page 2
http://mathlover.info/page/2
Newer posts ». Zoj3201 Tree of Tree. 分析 建好图 别忘了双向边 ，dp[u][j]表示以u为根的子树中选择体积为j的最大权。 递推式子为dp[u][j]=max(dp[u][j],dp[u][k] dp[v][j-k]);. 题意 有一个n*m(n,m =50)的格子，上面有不超过14根直线，每条直线至少经过3个点，然后现在告诉你(n 1)*(m 1)个格点每个格点上直线的条数，求最少直线数。 Ret) { ans=min(ans,s); return; } for(int i=0;i n; i) { for(int j=0;j m; j) { if(cnt[i][j]) { if(i= 0) { bool flag=0; for(int k=0;k n; k) if(! Cnt[k][j]) { flag=1; break; } if(! Cnt[i][k]) { flag=1; break; } if(! Qempty() { int u=q.front(); q.pop(); vis[u]=0; for(int i=head[u]; i;i=e...Visy[...
2014 - 数一新天地
http://mathlover.info/mcs/contest/2014
2014-2015 ACM-ICPC, NEERC, Moscow Subregional Contest解题报告. 传送门 http:/ codeforces.com/gym/100519. 传送门 http:/ codeforces.com/gym/100519/problem/A. 题意 模拟一个2048的游戏，不过地图是N*N的(N =100)，然后一开始里面有K个方块，并告诉你方块的数字以及坐标，然后有L个操作(L =10000)，每个操作，首先是一个代表上U下D左L右R方向的字母，然后是新增方块的位置以及数字 2或者4。 每次合成两个相同的带着数字2 n的块，会变成一个2 (n 1)的块，并获得2 (n 1)分。 不过显然旋转的复杂度是O(N 2),那么整个程序的复杂度就是O(LN 2)，1e8*谜一样的常数，得亏是CF的机子够强悍 670msAC。 0) p[cnt ]=x[i][j]; } for(int j=0; j cnt; j) if(p[j]= p[j 1]) res =p[j]*2,p[j]*=2,p[j 1]=0,j ; int l=0; while(p[l]! Includ...
Moscow Subregional - 数一新天地
http://mathlover.info/mcs/contest/moscow-subregional
Category Archives: Moscow Subregional. 2014-2015 ACM-ICPC, NEERC, Moscow Subregional Contest解题报告. 传送门 http:/ codeforces.com/gym/100519. 传送门 http:/ codeforces.com/gym/100519/problem/A. 题意 模拟一个2048的游戏，不过地图是N*N的(N =100)，然后一开始里面有K个方块，并告诉你方块的数字以及坐标，然后有L个操作(L =10000)，每个操作，首先是一个代表上U下D左L右R方向的字母，然后是新增方块的位置以及数字 2或者4。 每次合成两个相同的带着数字2 n的块，会变成一个2 (n 1)的块，并获得2 (n 1)分。 不过显然旋转的复杂度是O(N 2),那么整个程序的复杂度就是O(LN 2)，1e8*谜一样的常数，得亏是CF的机子够强悍 670msAC。 0) l ; for(int j=l 1; j cnt; j) { if(p[j]! Include bits/stdc .h. Cout res endl;.
poj - 数一新天地
http://mathlover.info/mcs/online-judge/poj
分析 显然就是要求一个最小的a，使得(n-m)a%L=x-y,整理一下就是同余线性方程，扩展欧几里得即可 注意范围会超过int，所以要用long long。 分析 最小费用最大流，看成在1 n之间找两条路，因为每条边只能走一次，所以每条边的容量为1，费用为路程，因为是无向图，所以要连反向边，然后源点与1之间连一条容量为2，费用为0的边，同样的n与汇点也连一条容量为2，费用为0的边，总共(2 2m)*2条边，注意内存别开小了，跑一次最小费用最大流就行了，spfa mcmf。 Qempty() { int u=q.front(); q.pop(); vis[u]=0; for(int i=head[u]; i;i=edge[i].next) { int v=edge[i].to; if(edge[i].cap edge[i].flow& dis[v] dis[u] edge[i].cost) { dis[v]=dis[u] edge[i].cost; pre[v]=i; if(! Poj2151 Check the difficulty of problems.
TOTAL LINKS TO THIS WEBSITE
21
Artiea Capital Management, Westlake Village, CA
2659 Townsgate Road, Ste. 203. Markets at a Glance. Our goal is to develop a relationship with you in which we play a key role in your financial life. The first steps include identifying your risk tolerance and goals, prioritizing them, and crafting a plan to help you realize your vision for the future. Our team will provide ongoing monitoring and advice for your personalized plan while tracking your progress as you pursue your financial goal. Who Needs a Succession Plan? Focused on College Funding.
武汉大学ACM/ICPC协会——主页 - Program Our Future
例子，字符串长度为10 6时，那么长度为6的字符串在其中最多出现了10 6 - 7 1次，. 但长度为6的字符串最多有10 7 - 10 6个,一定会出现重复。 用一下等比数列求和公式，发现最后结果为(A X - 1) / (A - 1)，因为题目中给出. 的限制条件，gcd(A, M) = gcd(A-1, M) = 1，可以发现/(A - 1)可以处理成某个非0的. 于是考虑X取何值时，A X % M = 1。 可以用反证法，假设存在一个比phi(M)更小的Y，满足A Y % M = 1，且Y不能够整除. Phi(M)，那么相当于存在A (phi(M) % Y) = 1, 而phi(M) % Y又必然小于Y，那么. 而dp(v1), dp(v2), dp(v3)则分别表示v1, v2, v3以下所有拓扑关系全部满足时. 此时u显然是u这个子树中第一个出现在最后生成的子序列中的数，而v1, v2, v3则为. Dp[u] = dp[v1] * C( v1 1 v2 - 1, v2 ) * dp[v2] * C( v1 v2 v3 1 - 1, v3 ). 总方案数为C(n m - 1, m)。
WPI ACM
Tuesdays at 8:00 PM in Fuller Commons. 2013-2014 ACM Networking Dinner. Posted by Ian Naval. On December 11, 2013. Come enjoy dinner with representatives from AdHarmonics, Google, KAYAK, MathWorks, Microsoft, VistaPrint and Wayfair! The dinner is Thursday, January 30th, 2014 at 6pm in the Higgins House. There is a $10 entry fee for students. The networking dinner is a great opportunity to meet professionals in the CS industry. This year's networking dinner is sponsored by VistaPrint. Posted by Ian Naval.
WashU ACM
Welcome to WashU ACM. We are the student chapter of the Association for Computing Machinery. At Washington University in St. Louis. We sponsor coding challenges, presentations, colloquia, workshops, and social events in Computer Science and Engineering. We're open to all students, whether or not you are majoring in CSE. Join us on Portfolio. Subscribe to our mailing list. And follow us on Twitter. Helping WashU developers find ideas. Are you a CSE student looking for a cool project to make? Watch out for...
Welcome To Online Judge
Xidian ACM Online Judge. Http:/ acm.xidian.edu.cn/download/seat.xlsx. Http:/ acm.xidian.edu.cn/download/rule.doc. Http:/ acm.xidian.edu.cn/download/ranklist.xlsx. 负责人 12-电院-山水 贪心只能过样例 CF老是FST. Http:/ acm.xidian.edu.cn/download/editorial.pdf. Anything about the Problems, Please Contact Admin: admin.
安徽财经大学管理科学与工程学院
Middot; 附件2 安徽财经大学博士科研基金管理办. Middot; 附件1 安徽财经大学博士科研基金申请表. Middot; 附件1 安徽财经大学博士科研基金申请表. Middot; 附件1 安徽财经大学博士科研基金申请表. Middot; 附件 2015年安徽财经大学教学骨干暑. Middot; 管工学院2015届毕业论文 设计 替代. Middot; 管工学院2015届毕业论文 设计 替代. 地址 安徽省蚌埠市曹山路962号 邮编 233030.
Welcome To XMU JudgeOnline
欢迎访问 acm.xmu.edu.cn/JudgeOnline. Welcome to XMU JudgeOnline. Problem Set is the place where you can find large amount of problems from different programming contests. Online Judge System allows you to test your solution for every problem. First of all, read carefully Frequently Asked Questions. Then, choose problem. Solve it and submit. If you want to publish your problems or setup your own online contest, just write us:. Web Designer: Zheng Le. Judge Developer: Xiang Guangte.
Untitled Document
Welcome to ZJGSU Online Judge
Check every 10 minutes. ZJGSU Online Judge Version 15.03.01 @ Github.
