~Love() - 博客园

确定比赛名次

摘要：确定比赛名次Time Limit : 2000/1000ms (Java/Other)Memory Limit : 65536/32768K (Java/Other)Total Submission(s) : 4Accepted Submission(s) : 2Font:Times New Roman|Verdana|GeorgiaFont Size:←→Problem Description有N个比赛队（1 2 #include 3 int degree[505],vis[505],map[501][501]; 4 int main() 5 { 6 int n,m,a,b,i,j,... 阅读全文

posted @ 2013-10-16 20:01 ~Love() 阅读(665) 评论(0) 推荐(0)

回文字符串

摘要： 1 #include 2 3 #include 4 int a[27],vis[11],len,k,t,cnt; 5 char str[11],temp[6],b_str[6],jd_str[121][6],c; 6 int judge() 7 { 8 int i; 9 for(i = 0;i 0;j --)25 printf("%c",temp[j]);26 printf("%c\n",temp[0]);27 return ;28 }29 for(i = 0;i 1)64 ... 阅读全文

posted @ 2013-10-13 19:34 ~Love() 阅读(177) 评论(0) 推荐(0)

翻纸牌游戏

摘要：翻纸牌游戏Time Limit : 9000/3000ms (Java/Other)Memory Limit : 32768/32768K (Java/Other)Total Submission(s) : 6Accepted Submission(s) : 2Font:Times New Roman|Verdana|GeorgiaFont Size:←→Problem Description有一种纸牌游戏，很有意思，给你N张纸牌，一字排开，纸牌有正反两面，开始的纸牌可能是一种乱的状态（有些朝正，有些朝反），现在你需要整理这些纸牌。但是麻烦的是，每当你翻一张纸牌（由正翻到反，或者有反翻到正）时阅读全文

posted @ 2013-09-25 16:15 ~Love() 阅读(444) 评论(0) 推荐(0)

骨牌铺方格

摘要：骨牌铺方格Time Limit : 2000/1000ms (Java/Other)Memory Limit : 65536/32768K (Java/Other)Total Submission(s) : 4Accepted Submission(s) : 2Font:Times New Roman|Verdana|GeorgiaFont Size:←→Problem Description在2×n的一个长方形方格中,用一个1× 2的骨牌铺满方格,输入n ,输出铺放方案的总数.例如n=3时,为2× 3方格，骨牌的铺放方案有三种,如下图：Input输入数据由多行组阅读全文

posted @ 2013-09-25 14:04 ~Love() 阅读(159) 评论(0) 推荐(0)

How many ways

摘要： How many waysTime Limit: 3000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2121Accepted Submission(s): 1279Problem Description这是一个简单的生存游戏，你控制一个机器人从一个棋盘的起始点(1,1)走到棋盘的终点(n,m)。游戏的规则描述如下：1.机器人一开始在棋盘的起始点并有起始点所标有的能量。2.机器人只能向右或者向下走，并且每走一步消耗一单位能量。3.机器人不能在原地停留。4.当机器人选择了一阅读全文

posted @ 2013-09-25 13:38 ~Love() 阅读(168) 评论(0) 推荐(0)

HDOj-1016 Prime Ring Problem

摘要： Prime Ring ProblemTime Limit: 4000/2000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 21132Accepted Submission(s): 9457Problem DescriptionA ring is compose of n circles as shown in diagram. Put natural number 1, 2, ..., n into each circle separately, and the sum of nu 阅读全文

posted @ 2013-09-25 00:03 ~Love() 阅读(156) 评论(0) 推荐(0)

DHU-1241 Oil Deposits

摘要： Oil DepositsTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8515Accepted Submission(s): 4980Problem DescriptionThe GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSurvComp works with one large rectangul 阅读全文

posted @ 2013-09-24 21:32 ~Love() 阅读(136) 评论(0) 推荐(0)

Red and Black

摘要： Red and BlackTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6865Accepted Submission(s): 4359Problem DescriptionThere is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. Fro 阅读全文

posted @ 2013-09-24 18:04 ~Love() 阅读(176) 评论(0) 推荐(0)

HDU-3790 最短路径问题

摘要：最短路径问题Time Limit : 2000/1000ms (Java/Other)Memory Limit : 32768/32768K (Java/Other)Total Submission(s) : 10Accepted Submission(s) : 3Font:Times New Roman|Verdana|GeorgiaFont Size:←→Problem Description给你n个点，m条无向边，每条边都有长度d和花费p，给你起点s终点t，要求输出起点到终点的最短距离及其花费，如果最短距离有多条路线，则输出花费最少的。Input输入n,m，点的编号是1~n,然后是m行，阅读全文

posted @ 2013-09-24 14:09 ~Love() 阅读(160) 评论(0) 推荐(0)

vim/Gvim配置

摘要： " Sections:" -> General" -> VIM user interface" -> Colors and Fonts" -> Files and backups" -> Text, tab and indent related" -> Visual mode related" -> Moving around, tabs and buffers" -> Status line" -> Editing mappings&quo 阅读全文

posted @ 2013-09-22 23:16 ~Love() 阅读(656) 评论(0) 推荐(1)