Code Force 21B Intersection】的更多相关文章

B. Intersection time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output You are given two set of points. The first set is determined by the equation A1x + B1y + C1 = 0, and the second one is determined…
1.debug,全部打印 2.打断点debug,出现单步调试等按钮,只运行断点前 3.setup over 调试一行代码 4.setup out 运行断点后面所有代码 5.debug窗口显示调试按钮 6.运行到对应的点会显示变量的值 7.step into:单步执行,遇到子函数就进入并且继续单步执行(简而言之,进入子函数): step over:在单步执行时,在函数内遇到子函数时不会进入子函数内单步执行,而是将子函数整个执行完再停止,也就是把子函数整个作为一步.有一点,经过我们简单的调试,在不存…
题目链接 大致题意 把一个图分成三块,要求任意两块之间是完全图,块内部没有连线 分析 首先根据块内没有连线可以直接分成两块 假定点1是属于块1的,那么所有与点1连接的点,都不属于块1:反之则是块1的 然后在所有不属于块1的点内随意找一点k,设定其属于块2,那么所有与点k连接的点且不属于块1,则是块3. 块分完了,然后是判断每个块是否满足条件,我通过下面三条来判断 1.每个块都有点 2.每个块内部没有连线,即没有一条线的两个端点在同一个块内 3.每个块内的点的度等于其他两个块的点个数和也等于n减去…
There are n people and k keys on a straight line. Every person wants to get to the office which is located on the line as well. To do that, he needs to reach some point with a key, take the key and then go to the office. Once a key is taken by somebo…
C. Andryusha and Colored Balloons time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Andryusha goes through a park each day. The squares and paths between them look boring to Andryusha, so he…
B. The Meeting Place Cannot Be Changed time limit per test 5 seconds memory limit per test 256 megabytes input standard input output standard output The main road in Bytecity is a straight line from south to north. Conveniently, there are coordinates…
B. Game of Credit Cards time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output After the fourth season Sherlock and Moriary have realized the whole foolishness of the battle between them and deci…
Summer is coming! It's time for Iahub and Iahubina to work out, as they both want to look hot at the beach. The gym where they go is a matrix a with n lines and mcolumns. Let number a[i][j] represents the calories burned by performing workout at the…
算是一题普通数论+思维题吧. 大概很多人是被题意绕晕了. 思路: 首先常规操作求出X的质因子. 然后题目要求的是,X的每个质因子p,在g(i,p)的连乘.i∈[1,n]: 我们转换下思维,不求每一个g(i,p)中最终是哪些 p的幂次,而是反求 每个p的幂次对结果的贡献. 显而易见,p^k在1~n的出现的次数就是  [n/(p^k)]. 这样枚举所有质因子,计算中再利用快速幂取模便可以得到答案 //#pragma comment(linker, "/STACK:1024000000,1024000…
1.利用 .htaccess 防止盗链 如果不喜欢别人在他们的网页上链接自己的图片.文档的话,也可以通过htaccess的指令来做到.当然这样也可以对你的网站服务器压力变小! 这次先给出‘代码’,然后进行详细的讲解!这个东西纠结了我很久啊,既然自己懂一些,就拿出来和大家一起分享一下,可能有些地方理解错误了,希望发现不对的‘童鞋’能指出,共同进步! RewriteCond %{HTTP_REFERER} !^$ RewriteCond %{HTTP_REFERER} !^http://localh…