Descirbe You are policeman and you are playing a game with Slavik. The game is turn-based and each turn consists of two phases. During the first phase you make your move and during the second phase Slavik makes his move. There are n doors, the i-th d…

题目链接:Doors Breaking and Repairing 题目大意:有n个门,先手攻击力为x(摧毁),后手恢复力为y(恢复),输入每个门的初始“生命值”,当把门的生命值攻为0时,就无法恢复了.问:最多可以把几个门的生命值攻为0. 思路:(1)当 x>y 的时候肯定所有的门的生命值都能降为0: (2)当 x<=y 的时候,先手的最优策略就是每次去攻击那些当前“生命值”比自己攻击力小的门,使它们的生命值降为0: 后手的最优策略就是去提高那些“生命值”比先手小的门的“生命值”,来减少先手“…

You are policeman and you are playing a game with Slavik. The game is turn-based and each turn consists of two phases. During the first phase you make your move and during the second phase Slavik makes his move. There are nn doors, the ii-th door ini…

题意:有\(n\)扇门,你每次可以攻击某个门,使其hp减少\(x\)(\(\le 0\)后就不可修复了),之后警察会修复某个门,使其hp增加\(y\),问你最多可以破坏多少扇门? 题解:首先如果\(x>y\),那么我肯定全部都能破坏,否则,统计\(hp\le x\)的门的个数,谁先碰门谁先赢,而我是先手,所以能破坏的门的个数就是\(\lceil \frac{cnt}{2} \rceil\). 代码: int n,x,y; int a[N]; int main() { //ios::sync_wi…

Codeforces Round #531 (Div. 3) 题目总链接:https://codeforces.com/contest/1102 A. Integer Sequence Dividing 题意: 给一个数n,然后要求你把1,2.....n分为两个集合,使得两个集合里面元素的和的差的绝对值最小. 题解: 分析可以发现,当n%4==0 或者 n%3==0,答案为0:其余答案为1.之后输出一下就好了. 代码如下: #include <bits/stdc++.h> using name…

Reinvent the Wheel Often Jason P. Sage Just use something that exists-it's silly to reinvent the wheel-. HAVE YOU EVER HEARD THiS OR SOME VARiATiON THEREOF? Sure you have! Every developer and student probably hears comments like this fre- quently. Wh…

How can I protect derived classes from breaking when I change the internal parts of the base class? A class has two distinct interfaces for two distinct sets of clients: It has a public interface that serves unrelated classes It has a protected inter…

题目链接:http://poj.org/problem?id=1102 题目意思:就是根据给出的格式 s 和 数字 n,输出数值 n 的 LCD 显示.数值 n 的每个数字要占据 s + 2 列 和 2s + 3 行.数字和数字之间要有一个空格.数值与数值之间有一个空行. 首先对于LCD 的 7 个笔画显示编上序号 然后对于数字 i,分析出占用了哪几个笔画,例如,数字 1 占有的笔画是 3 和 6:数字 6 占有的笔画是 1, 2, 4, 5, 6, 7 用数组来存储每一个笔画分别被那些数字占有…

Bahosain is walking in a street of N blocks. Each block is either empty or has one lamp. If there is a lamp in a block, it will light it’s block and the direct adjacent blocks. For example, if there is a lamp at block 3, it will light the blocks 2, 3…

1.HDU  1102  Constructing Roads    最小生成树 2.总结: 题意:修路,裸题 (1)kruskal //kruskal #include<iostream> #include<cstring> #include<cmath> #include<queue> #include<algorithm> #include<cstdio> #define max(a,b) a>b?a:b using na…