CSU-2019 Fleecing the Raffle Description A tremendously exciting raffle is being held, with some tremendously exciting prizes being given out. All you have to do to have a chance of being a winner is to put a piece of paper with your name on it in th…

题目: A tremendously exciting raffle is being held, with some tremendously exciting prizes being given out. All you have to do to have a chance of being a winner is to put a piece of paper with your name on it in the raffle box. The lucky winners of th…

Description A tremendously exciting raffle is being held, with some tremendously exciting prizes being given out. All you have to do to have a chance of being a winner is to put a piece of paper with your name on it in the raffle box. The lucky winne…

A. Artwork 倒过来并查集维护即可. #include<cstdio> #include<algorithm> using namespace std; const int N=1111; int n,m,q,i,j,ce; bool black[N][N]; bool v[N*N]; int f[N*N]; int res; int ans[N*N],id[N][N],cnt; struct P{ int x,y; P(){} P(int _x,int _y){x=_x,…

A Artwork B Bless You Autocorrect! C Card Hand Sorting D Daydreaming Stockbroker 贪心,低买高卖,不要爆int. #include <cstdio> #include <cstring> #include <algorithm> #include <iostream> using namespace std; typedef long long LL; const int max…

近日,有报道表示台积电10nm 芯片可怜的收益率可能会对 2017 年多款高端移动设备的推出产生较大的影响,其中自然包括下一代 iPhone 和 iPad 机型.不过,台积电正式驳斥了这一说法,表明10纳米处理器生产完全按计划进行,并没有出现任何延滞,并且这款处理器在明年第一季度就会开始盈利. 同时台积电还放出另一个重磅消息.台积电还表示,明年公司的生产计划是7纳米处理器,到2019年会生产5nm处理器,支持高端移动设备获得新功能,如VR/AR等.今年10月也有报道显示台积电已经开始进行于201…

传送门:csu 1812: 三角形和矩形 思路:首先,求出三角形的在矩形区域的顶点,矩形在三角形区域的顶点.然后求出所有的交点.这些点构成一个凸包,求凸包面积就OK了. /************************************************************** Problem: User: youmi Language: C++ Result: Accepted Time: Memory: ***********************************…

题目链接:http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1503 解题报告:分两种情况就可以了,第一种是那个点跟圆心的连线在那段扇形的圆弧范围内,这样的话点到圆弧的最短距离就是点到圆心的距离减去半径然后再取绝对值就可以了,第二种情况是那个点跟圆心的连线不在那段扇形的圆弧范围内,这样的话,最短的距离就是到这段圆弧的端点的最小值. 接下来的第一步就是求圆心的坐标跟圆的半径,只要求出圆心坐标半径就好说了,求圆心坐标我用的方法是: 设圆心坐标是(a,b…

题目链接:http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1120 解题报告:dp,用一个串去更新另一个串,递推方程是: if(b[i] > a[j]) m = max(m,dp[j]); else if(b[i] == a[j]) dp[j] = m + 1; #include<cstdio> #include<cstring> #include<iostream> #include<algorithm&g…

题目链接:http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1116 解题报告:一个国家有n个城市,有m条路可以修,修每条路要一定的金币,现在这个国家只有K个金币,每个城市有一些人,要你求怎么修路使得总的费用在K的范围内,同时使得跟首都连接的城市的人口(包括首都的人口)要最多,问最多的人口是多少. 枚举连接哪些城市,然后分别构造最小生成树. #include<cstdio> #include<cstring> #include<…