Problem Description A bit string has odd parity if the number of 1's is odd. A bit string has even parity if the number of 1's is even.Zero is considered to be an even number, so a bit string with no 1's has even parity. Note that the number of 0's d…

Parity Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1855    Accepted Submission(s): 1447 Problem Description A bit string has odd parity if the number of 1's is odd. A bit string has even pa…

Parity Time Limit: 2000/1000 MS(Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4334    Accepted Submission(s): 3264 Problem Description A bit string hasodd parity if the number of 1's is odd. A bit string has even pari…

http://acm.hdu.edu.cn/showproblem.php?pid=2700 题目意思很重要:  //e:是要使字符串中1的个数变成偶数.o:是要使字符串中1的个数变成奇数 Parity Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1855    Accepted Submission(s): 1447 Proble…

HDU 模拟题, 枚举1002 1004 1013 1015 1017 1020 1022 1029 1031 1033 1034 1035 1036 1037 1039 1042 1047 1048 1049 1050 1057 1062 1063 1064 1070 1073 1075 1082 1083 1084 1088 1106 1107 1113 1117 1119 1128 1129 1144 1148 1157 1161 1170 1172 1177 1197 1200 1201…

1.POJ 1733 Parity game Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5744   Accepted: 2233 Description Now and then you play the following game with your friend. Your friend writes down a sequence consisting of zeroes and ones. You cho…

转自——http://blog.csdn.net/qwe20060514/article/details/8112550 =============================以下是最小生成树+并查集======================================[HDU]1213   How Many Tables   基础并查集★1272   小希的迷宫   基础并查集★1325&&poj1308  Is It A Tree?   基础并查集★1856   More i…

转载来自:http://www.cppblog.com/acronix/archive/2010/09/24/127536.aspx 分类一: 基础题:1000.1001.1004.1005.1008.1012.1013.1014.1017.1019.1021.1028.1029.1032.1037.1040.1048.1056.1058.1061.1070.1076.1089.1090.1091.1092.1093.1094.1095.1096.1097.1098.1106.1108.1157…

模拟题, 枚举1002 1004 1013 1015 1017 1020 1022 1029 1031 1033 1034 1035 1036 1037 1039 1042 1047 1048 1049 1050 1057 1062 1063 1064 1070 1073 1075 1082 1083 1084 1088 1106 1107 1113 1117 1119 1128 1129 1144 1148 1157 1161 1170 1172 1177 1197 1200 1201 120…

一.思路很简单,搜索.对于每一种状态,利用康托展开编码成一个整数.于是,状态就可以记忆了. 二.在搜索之前,可以先做个优化,对于逆序数为奇数的序列,一定无解. 三.搜索方法有很多. 1.最普通的:深搜.广搜.在这题里面,这两个方法直接TLE.所以,我后面没有贴超时的代码. 2.既然1超时,那就预处理出所有状态,用map存储,然而,map的insert(使用[]是一样的)实在太慢了,也超时. 3.在1的基础上,优化一下,得到:IDA*,双向广搜,A*. 3.IDA*我没尝试,不过感觉没有A*快.另…