英文字典中文字典


英文字典中文字典51ZiDian.com



中文字典辞典   英文字典 a   b   c   d   e   f   g   h   i   j   k   l   m   n   o   p   q   r   s   t   u   v   w   x   y   z       







请输入英文单字，中文词皆可：


请选择你想看的字典辞典：
单词字典翻译
scateren查看　scateren　在百度字典中的解释百度英翻中〔查看〕
scateren查看　scateren　在Google字典中的解释Google英翻中〔查看〕
scateren查看　scateren　在Yahoo字典中的解释Yahoo英翻中〔查看〕

安装中文字典英文字典查询工具!


中文字典英文字典工具:
选择颜色:
输入中英文单字

































































英文字典中文字典相关资料:


  • Codeforces
    Hello Codeforces! We are really excited to invite you to participate in Codeforces Round 1089 (Div 2), which will be held on Sunday, March 29, 2026 at 07:35 UTC-7 You will be given 6 6 "simple" problems and 2 2 hour 15 15 minutes to solve them Note that one of the problems will be further divided into 2 2 subtasks Furthermore, some of the problems may be interactive, so please read the
  • Problemset - Codeforces
    Codeforces Programming competitions and contests, programming community
  • login - Codeforces
    Codeforces Programming competitions and contests, programming community Contest rating: 2154 (max master, 2154) Contribution: 0 Friend of: 770 users Last visit: 4 years ago Registered: 9 years ago Comments Talks | Send message
  • Login - Codeforces
    Codeforces Programming competitions and contests, programming community Server time: Apr 05 2026 07:17:54 (f1)
  • Codeforces
    Hello Codeforces! We are really excited to invite you to participate in Codeforces Round 1089 (Div 2), which will be held on Sunday, March 29, 2026 at 07:35 UTC-7 You will be given 6 6 "simple" problems and 2 2 hour 15 15 minutes to solve them Note that one of the problems will be further divided into 2 2 subtasks Furthermore, some of the problems may be interactive, so please read the
  • Login - Codeforces
    Fill in the form to login into Codeforces Login into Codeforces Handle Email Password Remember me for a month
  • Codeforces Round 1079 (Div. 1, Div. 2) Editorial - Codeforces
    Codeforces Programming competitions and contests, programming community Consider how to determine if a sequence is valid Maintain two sums x, y x, y For (, x ← x + 1 x ← x + 1; for ), y ← y + 1 y ← y + 1, and square brackets do the similar operation on y y A sequence is valid if and only if x, y ≥ 0 x, y ≥ 0 at any moment, and finally x = y = 0 x = y = 0 For the single bracket
  • Catalog - Codeforces
    Codeforces Programming competitions and contests, programming community 3 months ago User LeoPro inserted blog post Maths for mincut problem: do we really need Dinic to get mincut? maths for mincut problem:Do we really need to use Dinic to get mincut? by OtterZ: inserted field blogEntryId 146966, inserted field catalogFolder Flows, inserted field creationTime Dec 22, 2025, 12:35:02 PM
  • Rule Restricting the Use of AI [revision 2024-09-14] - Codeforces
    This includes both registered contestants and out-of-competition participants in any rated Codeforces round If a round is unrated and this is explicitly stated in the round announcement or rules, the rule will also apply Outside of such contests, AI tools may be used freely for practice, learning, or non-competitive problem-solving
  • Top - Codeforces
    Codeforces Programming competitions and contests, programming community Hello codeforces, welcome back to another cheater callout post Today, we have user Jochen_Solve While this user is already getting skipped and further action may not be necessary, I decided to put this user to the test on my own After asking him to solve a problem I put in a mashup contest and edited, he left some





中文字典-英文字典  2005-2009