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…

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…

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…

After the fourth season Sherlock and Moriary have realized the whole foolishness of the battle between them and decided to continue their competitions in peaceful game of Credit Cards. Rules of this game are simple: each player bring his favourite n-…

After the fourth season Sherlock and Moriary have realized the whole foolishness of the battle between them and decided to continue their competitions in peaceful game of Credit Cards. Rules of this game are simple: each player bring his favourite n-…

After the fourth season Sherlock and Moriary have realized the whole foolishness of the battle between them and decided to continue their competitions in peaceful game of Credit Cards. Rules of this game are simple: each player bring his favourite \(…

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…

本文链接:https://www.cnblogs.com/blowhail/p/10990833.html Nauuo and Cards 原题链接:http://codeforces.com/contest/1173/problem/C 题目大意 :有2n张卡片,其中n张编号为1-n,另外n张为0. 现在手中拿n张,桌子上放n张,每次操作可以在桌子上底部放入一张卡片,最上面拿走一张卡片,问最小操作次数. 思路: 先看看卡片1的位置,卡片1在桌子上时,看看是否满足从1递增到底部,例如 0 0 1…

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…