B - Magnets】的更多相关文章

C. Edo and Magnets Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/594/problem/C Description Edo has got a collection of n refrigerator magnets! He decided to buy a refrigerator and hang the magnets on the door. The shop ca…
Problem description Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If…
CodeForces - 344A id=46664" style="color:blue; text-decoration:none">Magnets Time Limit: 1000MS   Memory Limit: 262144KB   64bit IO Format: %I64d & %I64u Submit Status Description Mad scientist Mike entertains himself by arranging ro…
Description Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two mag…
传送门:D. Monopole Magnets 这一场也是很神奇了,先是推迟三天,后是评测鸡崩了,unrated... 题意:每一行,每一列必须都要至少有一个s,n要可以到所有的黑格,n的上下左右如果有一个s,他就可以往那个方向移动一格,n最后不能在白格,求n最少的个数. 题解: 1.每一行每一列如果有#,#必须是连续的,否则输出-1 2.如果有一行没有#,那么至少有一列要没有#,s就可以放在交点处,否则输出-1. 3.求解连通块的个数 1 #include<bits/stdc++.h> 2…
link:http://codeforces.com/contest/344/problem/A 这道题目很简单. 把输入的01 和10 当做整数,如果相邻两个数字相等的话,那么就属于同一组,否则,就新增加了一组. #include <cstdio> #include <cstdlib> #include <cstring> using namespace std; int main(void) { #ifndef ONLINE_JUDGE freopen("…
首先,只需要找到一个有磁性的位置,就可以通过$n-1$次判断其余磁铁是否有磁性,因此也就是要在$\lfloor\log_{2}n\rfloor+1$次中找到一个有磁性的位置 有一个$n-1$次的做法,即暴力枚举第$i$个磁铁($i\ge 2$),将1到$i-1$的磁铁放在左侧,那么一定能找到第2个有磁性的磁铁,由于总存在两个,即可以找到 事实上,找磁铁已经无法优化了,但找磁铁的过程却可以带来额外的信息:假设第一个磁铁位于$i$,$i$之前恰好存在一个磁铁,对$i$之前的部分二分即可 更精确的来说…
这次比赛出的题真是前所未有的水!只用了一小时零十分钟就过了前4道题,不过E题还是没有在比赛时做出来,今天上午我又把E题做了一遍,发现其实也很水.昨天晚上人品爆发,居然排到Rank 55,运气好的话没准能领到T-shirt.除此之外,锁上程序之后,看到一个人数组开小了,我还提交了一个大数据,成功Hack了一次,然后Room排名顿时升到第1. My submissions     # When Who Problem Lang Verdict Time Memory 4474604 Sep 15,…
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