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原创 杭州邀请赛Robot
DescriptionMichael has a telecontrol robot. One day he put the robot on a loop with n cells. The cells are numbered from 1 to n clockwise. At first the robot is in cell 1. Then Michael u
2014-07-29 10:35:10
664
转载 高校联合Rating
由于每次增加分数或者是减少分数都是50的倍数,因而我们可以压缩成每次赢了增加一分,输了减少2分。根据题意我们容易看出,每次分数的变化都是最小的分数进行变化的。因而我们定义状态ans[i][j]表示从初状态到两个号的分数为i,j的期望。我们可以知道两个号的分数的变化总是[i,i]->[i+1,i]->[i+1,i+1];每次只有一个号的分数在进行变动,这样子就比较容易进行状态的转移。定义dp
2014-07-23 17:10:31
490
原创 hdu多校联合Peter's Hobby
Problem DescriptionRecently, Peter likes to measure the humidity of leaves. He recorded a leaf humidity every day. There are four types of leaves wetness: Dry , Dryish , Damp and Soggy. As we know
2014-07-23 09:44:46
536
原创 Polyomino Composer
A polyomino is a plane geometric figure formed by joining one or more equal squares edge to edge.- WikipediaGiven a large polyomino and a small polyomino, your task is to determine whether you
2014-07-19 10:07:02
685
原创 2014西安邀请赛部分题解
这道题很简单,最短路径的变形,只不过图的邻接矩阵(C【i】【j】)要自己先求出来,公式题目也给出来了。最短路径3种方法都能用,WA了几次全因为题目要求的结果没明白……读题真是硬伤啊。
2014-07-09 09:50:33
641
原创 有向图的邻接表建立及其深搜宽搜
#include#include#include#include#includeusing namespace std;typedef struct edge{ int no; struct edge *next;}edgetype;//链表节点typedef struct{ int mark; char vertex[10]; edge
2013-11-10 20:42:17
836
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