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摘要: Problem ATime Limit : 2000/1000ms (Java/Other)Memory Limit : 131072/65536K (Java/Other)Total Submission(s) : 9Accepted Submission(s) : 7Problem DescriptionA checksum is an algorithm that scans a packet of data and returns a single number. The idea is that if the packet is changed, the checksum will. 阅读全文
posted @ 2013-12-11 19:35 疯狂的癫子 阅读(148) 评论(0) 推荐(0) 编辑
摘要: http://acm.hdu.edu.cn/showproblem.php?pid=1035 Robot MotionTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5591Accepted Submission(s): 2604Problem DescriptionA robot has been programmed to follow the instructions in its path. Instructions for the n 阅读全文
posted @ 2013-12-11 19:30 疯狂的癫子 阅读(203) 评论(0) 推荐(0) 编辑
摘要: http://poj.org/problem?id=1782Run Length EncodingTime Limit:1000MSMemory Limit:30000KTotal Submissions:3907Accepted:1282DescriptionYour task is to write a program that performs a simple form of run-length encoding, as described by the rules below.Any sequence of between 2 to 9 identical characters i 阅读全文
posted @ 2013-12-02 20:27 疯狂的癫子 阅读(319) 评论(0) 推荐(0) 编辑
摘要: Problem ETime Limit : 4000/2000ms (Java/Other)Memory Limit : 32768/16384K (Java/Other)Total Submission(s) : 15Accepted Submission(s) : 7Problem DescriptionConsider the sequence of numbers ai, i = 0, 1, 2, …, which satisfies the following requirements:a0 = 0a1 = 1a2i = aia2i+1 = ai + ai+1for every i. 阅读全文
posted @ 2013-11-30 22:23 疯狂的癫子 阅读(153) 评论(0) 推荐(0) 编辑
摘要: Problem CTime Limit : 2000/1000ms (Java/Other)Memory Limit : 20000/10000K (Java/Other)Total Submission(s) : 9Accepted Submission(s) : 8Problem DescriptionA certain prison contains a long hall of n cells, each right next to each other. Each cell has a prisoner in it, and each cell is locked. One nig. 阅读全文
posted @ 2013-11-30 22:19 疯狂的癫子 阅读(209) 评论(0) 推荐(0) 编辑
摘要: Problem BTime Limit : 2000/1000ms (Java/Other)Memory Limit : 20000/10000K (Java/Other)Total Submission(s) : 10Accepted Submission(s) : 7Problem Description把M个同样的苹果放在N个同样的盘子里,允许有的盘子空着不放,问共有多少种不同的分法?(用K表示)5,1,1和1,5,1 是同一种分法。Input第一行是测试数据的数目t(0 int find(int m,int n) { if(m==0||n==1) ret... 阅读全文
posted @ 2013-11-30 22:17 疯狂的癫子 阅读(159) 评论(0) 推荐(0) 编辑
摘要: 细节输入的数据要特别考虑清楚。 A strange liftTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 9356Accepted Submission(s): 3548 Problem DescriptionThere is a strange lift.The lift can stop can at every floor as you want, and there is a number Ki(0 #include#define 阅读全文
posted @ 2013-11-29 22:16 疯狂的癫子 阅读(237) 评论(0) 推荐(0) 编辑
摘要: 最短路径问题Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 9352Accepted Submission(s): 2852 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,每行4个数 a,b,d,p,表示a和b之间有一条边,且其长度为d,花费为p。最后. 阅读全文
posted @ 2013-11-27 20:21 疯狂的癫子 阅读(185) 评论(0) 推荐(0) 编辑
摘要: 过河问题 没有找2个方法比较,只用了一个方法。wa了。 全局变量换成局部变量也wa了。换成memset(a,0,sizeof(a));就可以。/*题意: 有n个人过河,只有一条船,每次只能坐2人,每个人过河所需时间不同,求全部过河所需时间。解题思路:当人数等于1时:sum=Num[1]; 当人数等于2时:sum=Num[2]; 当人数等于3时:sum=Num[3]+Num[2]+Num[1];当人数大于等于4时: 若设过桥速度最快的那个人过桥时间为a,第二快为b;过桥第二慢的那个人过桥时间为c,最慢为d; 此时有两种过桥方案: 一.最快和次快的人先过,然后最快的回来,然后最慢与次慢的人再过,次 阅读全文
posted @ 2013-11-26 20:26 疯狂的癫子 阅读(261) 评论(0) 推荐(0) 编辑
摘要: http://acm.hdu.edu.cn/showproblem.php?pid=2522学习://除数的运算的应用和算法。除法的本质,如果余数出现重复就表示有循环节 A simple problemTime Limit: 4000/2000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2795Accepted Submission(s): 984 Problem DescriptionZty很痴迷数学问题.。一天,yifenfei出了个数学题想难倒他,让他回答1 / n。但Zt 阅读全文
posted @ 2013-11-25 20:57 疯狂的癫子 阅读(175) 评论(0) 推荐(0) 编辑
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