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Modified UTF-8 �?UTF-32 �怺�转换

coreBugZJ — Sun, 13 Apr 2014 11:42:00 GMT

摘要: 自己的实玎ͼ��l�过一定的��试�?nbsp; 阅读全文

coreBugZJ 2014-04-13 19:42 发表评论

JNI Java调用C代码

coreBugZJ — Sun, 13 Apr 2014 11:36:00 GMT

摘要: 环境 Ubuntu12.04 + openJDK6�?
JNI中��用“修改的UTF8”编码，而Linux下wchar_t使用UTF32�Q�注意�{码�?nbsp; 阅读全文

coreBugZJ 2014-04-13 19:36 发表评论

微��Y2014实习(f��n)生及(qi��ng)�U��o(h��)营技术类职位在线��试

coreBugZJ — Sun, 13 Apr 2014 11:11:00 GMT

摘要: 阅读全文

coreBugZJ 2014-04-13 19:11 发表评论

COM �~�程获取墙纸路径

coreBugZJ — Mon, 18 Nov 2013 12:53:00 GMT

摘要: 初学COM�~�程�?nbsp; 阅读全文

coreBugZJ 2013-11-18 20:53 发表评论

LZW �~�码解码代码

coreBugZJ — Mon, 04 Nov 2013 07:58:00 GMT

摘要: �?libtiff 4.0.2 中提取出来�ƈ�E�加修改�?lzw 的代码，�W�合 TIFF6 标准中的 LZW 部分�?
本�h目前对开源协议还?sh��)��太清楚�Q�不知是否存在��R权问题，如果有，请告知�?nbsp; 阅读全文

coreBugZJ 2013-11-04 15:58 发表评论

coreBugZJ — Wed, 09 Oct 2013 12:13:00 GMT

coreBugZJ 2013-10-09 20:13 发表评论

coreBugZJ — Sat, 28 Sep 2013 09:03:00 GMT

问题来自 TopCoder SRM 591 DIV 2 的第二题�Q?br />
Problem Statement

Let X and Y be two strings of equal length, consisting of uppercase English letters only. The two strings are called convertible if there is a permutation P of the English alphabet with the following property: if each character c in the string X is replaced by the character P(c), we obtain the string Y. (In other words, X and Y are convertible iff the following holds: whenever two letters of X are equal, the corresponding two letters of Y must be equal, and vice versa.) For example, consider the string "ABCA". We can choose to replace each 'A' by a 'F', each 'B' by a 'B', and each 'C' by a 'G', obtaining the string "FBGF". Thus the strings "ABCA" and "FBGF" are convertible. The strings "ABCA" and "EFGH" are not convertible, because the two 'A's in the first string must correspond to the same letter in the second string. The strings "ABCA" and "EEEE" are not convertible, because different letters in the first string must correspond to different letters in the second string. You are given two strings A and B of the same length. These strings only contain English letters from 'A' to 'I', inclusive. (That is, only the first 9 letters of the alphabet are used.) You want to change A and B into two strings that are convertible. The only allowed change is to choose some indices (possibly none) and to remove the characters at those indices from each of the strings. (I.e., the removed characters must be at the same positions in both strings. For example, it is not allowed to only remove character 0 of A and character 3 of B.) For example, if A="ABAC", B="AHHA" and the chosen indices are 0 and 2, then the resulting strings will be "BC" and "HA". Our goal is to choose as few indices as possible, given that after the erasing we want to obtain two convertible strings. Compute and return the smallest possible number of chosen indices.
Definition

Class:
ConvertibleStrings
Method:
leastRemovals
Parameters:
string, string
Returns:
int
Method signature:
int leastRemovals(string A, string B)
(be sure your method is public)


Constraints
-
A will contain between 1 and 50 characters, inclusive.
-
A and B will be of the same length.
-
A will contain characters from 'A' to 'I' only.
-
B will contain characters from 'A' to 'I' only.

我的思�\是穷举A中字母与B中字母的对应关系�Q�看哪种对应需要删除的字母最��，�q�一最��值即是答案�?br />�I��D对应关系�Q�就是生成全排列�?br />我生成全排列的方式是回溯�?br />
之后看其他�h的代码，发现一个由�l�定排列求出其下一个排列的函数�Q�于是学�?f��n)一下，自己实现如下�Q?br />

// 生成下一字典序排�?br />// 假设a中元素互不相�?br />// 若已�l�是最后一个字典序排列�Q�则�q�回0�Q�否则返�?
int next_permutation( int a[], int n ) {
        int i, j;
        for ( i = n-1; (0 < i) && (a[i-1] > a[i]); --i ) {
        }
        if ( 0 >= i ) {
                return 0;
        }
        for ( j = n-1; j >= i; --j ) {
                if ( a[ j ] > a[ i-1 ] ) {
                        int tmp = a[ i-1 ];
                        a[ i-1 ] = a[ j ];
                        a[ j ] = tmp;
                        j = n - 1;
                        while ( i < j ) {
                                tmp = a[ i ];
                                a[ i ] = a[ j ];
                                a[ j ] = tmp;
                                ++i; --j;
                        }
                        break;
                }
        }
        return 1;
}

�q�有��Z��用的是C�Q�＋�? �?next_permutation 函数�Q�功能一栗��?br />

coreBugZJ 2013-09-28 17:03 发表评论

nasm x86 32位汇�~�实�?int64 带符�?�?�?�?�?比较大小

coreBugZJ — Mon, 29 Apr 2013 12:46:00 GMT

摘要: 毕业论文�?Pascal �~�译器，需要生成的32�?nasm 汇编代码支持 64位带�W�号整数阅读全文

coreBugZJ 2013-04-29 20:46 发表评论

汇编调用 C 语言 printf �Q�f(xi��) 输出��点�?float 的问题：(x��)4字节 �q�是 8字节

coreBugZJ — Mon, 29 Apr 2013 08:51:00 GMT

摘要: �l�论�Q?C 语言�?printf 使用 %f 来输�?float �Ӟ��实际上先�?4字节的float转化�?8字节的double�Q�然后访问了(ji��n)栈上�?8字节数据�?
�Q�环境：(x��)Ubuntu12.04 32�?intel i3 nasm gcc�Q?nbsp; 阅读全文

coreBugZJ 2013-04-29 16:51 发表评论

Linux �?Java 调用 shell 命��o(h��) nasm �?gcc

coreBugZJ — Mon, 29 Apr 2013 07:28:00 GMT

摘要: 最�q�在写Pascal�~�译器，生成 nasm 汇编代码�Q�需要调�?nasm �q�行汇编�Q�再调用 gcc �?C 语言标准库链接，从而可以��?C 语言的库函数�?

注意 "-o" �?fileName 要分开��Z��个字�W�串�Q�不然的话，nasm 没问题，�?gcc �?x��)出错，链接器找不到输出文�g�?

errReader �?errInfo 用于输出 nasm �?gcc 输出的错误信息�?nbsp; 阅读全文

coreBugZJ 2013-04-29 15:28 发表评论

coreBugZJ — Fri, 05 Apr 2013 09:05:00 GMT

摘要: �q�篇文章里主要阐�q�C��面几个问�? 1: 字符和字节的区别 2: Big Endian和Little Endian 3: ASCII 4: Unicode的定�? 5: Unicode的编码方�? 6: Unicode的实现方�? 7: 一些名词解�?nbsp; 阅读全文

coreBugZJ 2013-04-05 17:05 发表评论

coreBugZJ — Fri, 05 Apr 2013 08:48:00 GMT

摘要: 之前因�ؓ(f��)考研的事情，攑ּ��?ji��n)更斎ͼ�保研后也没回来。这一丢就是好久�?nbsp; 阅读全文

coreBugZJ 2013-04-05 16:48 发表评论

coreBugZJ — Wed, 04 Jul 2012 02:41:00 GMT

摘要: �?920�q�快要结束的一天，多��u多大学生理学�pȝ��麦克力_�d�Q�John James Rickard Macleod�Q�教授迎来了(ji��n)一位年��M�h的到�ѝ��来��辞恳切，声音略有些激动，但看上去游说�q�不是他的专�ѝ��在�q�轻��明来意之后，麦克力_�d教授�q�不为所动，而是以“需要考虑”�ؓ(f��)由婉拒了(ji��n)来�h的请求。看到教授如此态度�Q�年��M�h难掩��p��的神情，待了(ji��n)没多久就��d��?ji��n)。。�?nbsp; 阅读全文

coreBugZJ 2012-07-04 10:41 发表评论

coreBugZJ — Tue, 26 Jun 2012 06:22:00 GMT

摘要: ��d��的一��研�I�引起了(ji��n)我们的注意。传�l�观念认为，人的智商在幼儿期��已定型�Q�此后要��x(ch��ng)��变非常困难；但此��研�I�表明，智商不仅可以提升�Q�而且�q��(sh��)��是一两分的小数目。。�?nbsp; 阅读全文

coreBugZJ 2012-06-26 14:22 发表评论

coreBugZJ — Sun, 24 Jun 2012 16:07:00 GMT

摘要: 天才是什么？天才其实�q�没有一个详��完整的定义�Q�不像“多�q�生矮小灌木”或者“年度优�U��?x��)计”那样可以逐条列出达标条款。。�?nbsp; 阅读全文

coreBugZJ 2012-06-25 00:07 发表评论

coreBugZJ — Sun, 24 Jun 2012 14:44:00 GMT

摘要: 囄�� 阅读全文

coreBugZJ 2012-06-24 22:44 发表评论

coreBugZJ — Sun, 24 Jun 2012 14:35:00 GMT

摘要: 文化的不同导致了(ji��n)思维方式的不同，思维方式的不同导致了(ji��n)游戏规则的不同，游戏规则的不同导致了(ji��n)��王的含恨落败。。�?nbsp; 阅读全文

coreBugZJ 2012-06-24 22:35 发表评论

A* ��法求解八数码问题，POJ 1077 Eight

coreBugZJ — Tue, 05 Jun 2012 07:06:00 GMT

摘要: �l�典不解释。。�?nbsp; 阅读全文

coreBugZJ 2012-06-05 15:06 发表评论

k-means ��法实现人口聚类

coreBugZJ — Tue, 05 Jun 2012 07:04:00 GMT

摘要: �l�典��法�Q�第一�ơ尝试聚�c�R��。�?nbsp; 阅读全文

coreBugZJ 2012-06-05 15:04 发表评论

coreBugZJ — Tue, 05 Jun 2012 07:02:00 GMT

摘要: �l�典��法�Q�被我实现的极其恶心(j��)。。�?nbsp; 阅读全文

coreBugZJ 2012-06-05 15:02 发表评论

coreBugZJ — Mon, 04 Jun 2012 08:05:00 GMT

摘要: 博弈论的问题�Q�需要证明一些结论。。�?nbsp; 阅读全文

coreBugZJ 2012-06-04 16:05 发表评论

POJ 2068 Nim

coreBugZJ — Mon, 04 Jun 2012 08:03:00 GMT

摘要: 博弈DP �Q�记忆化搜烦(ch��)。。�?nbsp; 阅读全文

coreBugZJ 2012-06-04 16:03 发表评论

POJ 2975 Nim

coreBugZJ — Mon, 04 Jun 2012 08:01:00 GMT

摘要: �l�典 Nim 博弈。。�?nbsp; 阅读全文

coreBugZJ 2012-06-04 16:01 发表评论

coreBugZJ — Mon, 04 Jun 2012 07:25:00 GMT

摘要: 我们一直在喊敏捷开发，其实敏捷开发的一个很重要的目的就是消除浪费，防止破窗效应的发生。事情太难，��p��它简单，更简单。流�E�太重，��p��它轻点，更轻炏V��尽量扫清开发的�?��，消灭破窗形成的环境。下面我�?x��)从软�g构徏的很多方面来描述如何防止“��Y件开发中的破�H�”。。�?nbsp; 阅读全文

coreBugZJ 2012-06-04 15:25 发表评论

coreBugZJ — Mon, 04 Jun 2012 07:22:00 GMT

摘要: ��C��商务依靠多�Q务来完成工作。评价员工也��Z��的他们多��d��能力。IT业�h员�(sh��)��(x��)被例行指�z�ֈ�多个��目中去。我们是�l�常在这样做吗？多�Q务�v作用吗？多�Q务的真正影响是什么？有别的选择吗？。。�?nbsp; 阅读全文

coreBugZJ 2012-06-04 15:22 发表评论