前进 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5001    Accepted Submission(s): 3902 Problem Description 轻松通过墓碑,进入古墓后,才发现里面别有洞天.突然,Yifenfei发现自己周围是黑压压的一群蝙蝠,个个扇动翅膀正准备一起向他发起进攻!形势十分危急!好在此时的yifenf…

Problem Description 轻松通过墓碑,进入古墓后,才发现里面别有洞天. 突然,Yifenfei发现自己周围是黑压压的一群蝙蝠,个个扇动翅膀正准备一起向他发起进攻! 形势十分危急! 好在此时的yifenfei已经不是以前那个经常被lemon抢走MM的菜鸟了!面对众多蝙蝠的嗜血狂攻,只见yifenfei使出轻灵的剑法,刷,刷,刷,瞬间搞定-- 现已知yifenfei使用了2招(剑招A和剑招B):剑招A,一招能杀死一半的蝙蝠.但是如果当前的蝙蝠数为奇数,那么就必须先出一招剑招B杀死其中…

HDOJ 题目分类 //分类不是绝对的 //"*" 表示好题,需要多次回味 //"?"表示结论是正确的,但还停留在模块阶 段,需要理解,证明. //简单题看到就可以敲的 1000:    入门用: 1001:    用高斯求和公式要防溢出 1004:1012: 1013:    对9取余好了 1017:1021: 1027:    用STL中的next_permutation() 1029:1032:1037:1039:1040:1056:1064:1065: 10…

HDOJ 题目分类 /* * 一:简单题 */ 1000:    入门用:1001:    用高斯求和公式要防溢出1004:1012:1013:    对9取余好了1017:1021:1027:    用STL中的next_permutation()1029:1032:1037:1039:1040:1056:1064:1065:1076:    闰年 1084:1085:1089,1090,1091,1092,1093,1094, 1095, 1096:全是A+B1108:1157:1196:1…

FatMouse' Trade Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 56784    Accepted Submission(s): 19009 Problem Description FatMouse prepared M pounds of cat food, ready to trade with the cats g…

Nasty Hacks Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3049    Accepted Submission(s): 2364 Problem Description You are the CEO of Nasty Hacks Inc., a company that creates small pieces of…

Box of Bricks Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5994    Accepted Submission(s): 2599 Problem Description Little Bob likes playing with his box of bricks. He puts the bricks one up…

Problem Description Contest time again! How excited it is to see balloons floating around. But to tell you a secret, the judges' favorite time is guessing the most popular problem. When the contest is over, they will count the balloons of each color…

卧槽....最近刷的cf上有最短路,本来想拿这题复习一下.... 题意就是在输出最短路的情况下,经过每个节点会增加税收,另外要字典序输出,注意a到b和b到a的权值不同 然后就是处理字典序的问题,当松弛时发现相同值的时候,判断两条路径的字典序 代码 #include "stdio.h" const int MAXN=110; const int INF=10000000; bool vis[MAXN]; int pre[MAXN]; int cost[MAXN][MAXN],lowcos…

Rectangles    HDOJ(2056) http://acm.hdu.edu.cn/showproblem.php?pid=2056 题目描述:给2条线段,分别构成2个矩形,求2个矩形相交面积. 算法:先用快速排斥判断2个矩形是否相交.若不相交,面积为0.若相交,将x坐标排序去中间2个值之差,y坐标也一样.最后将2个差相乘得到最后结果. 这题是我大一的时候做过的,当时一看觉得很水,写起来发现其实没我想的那么水.分了好几类情况没做出来.今天看了点关于判断线段相交的知识,想起了这题便拿来练…