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微��Y2014实习生及�U��o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

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目前对开源协议还不太清楚�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�于是学习一下，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

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。。�?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。。�?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

POJ 3696 The Luckiest number

coreBugZJ — Fri, 01 Jun 2012 13:32:00 GMT

摘要: �l�一个数N(1<=N<=2000000000);问是否存在N的倍数M�Q�且M的各个位全部�?�l�成�Q�如果存在多个取最��的 M �q�输出M由几�?�l�成。。�?nbsp; 阅读全文

coreBugZJ 2012-06-01 21:32 发表评论

POJ 3604 Professor Ben

coreBugZJ — Fri, 01 Jun 2012 13:30:00 GMT

摘要: 求整数的所有的因子的因子数的立方和。。�?nbsp; 阅读全文

coreBugZJ 2012-06-01 21:30 发表评论

EOJ 1117 剩余定理

coreBugZJ — Fri, 01 Jun 2012 13:27:00 GMT

摘要: 求正整数中满��I��X mod a[0] = b[0], X mod a[1] = b[1], X mod a[2] = b[2], �? X mod a[i] = b[i], �?的最��解。a[i]是一些两两互质的正整数。。�?nbsp; 阅读全文

coreBugZJ 2012-06-01 21:27 发表评论

coreBugZJ — Fri, 01 Jun 2012 13:26:00 GMT

摘要: 形如ax≡b(mod m) 的方�E�，�U�Cؓ�U�性同余方�E�。编写程序求解线性同余方�E�（��Z��Ƨ几里�d��法�Q�。。�?nbsp; 阅读全文

coreBugZJ 2012-06-01 21:26 发表评论

POJ 1603 Risk

coreBugZJ — Mon, 14 May 2012 08:13:00 GMT

摘要: 图论最短�\�?nbsp; 阅读全文

coreBugZJ 2012-05-14 16:13 发表评论

EOJ 1848 你是ACM吗？

coreBugZJ — Mon, 14 May 2012 08:12:00 GMT

摘要: 图论最短�\�?nbsp; 阅读全文

coreBugZJ 2012-05-14 16:12 发表评论

coreBugZJ — Mon, 14 May 2012 08:08:00 GMT

摘要: 图论最短�\�?nbsp; 阅读全文

coreBugZJ 2012-05-14 16:08 发表评论

EOJ 1708 Connected Gheeves

coreBugZJ — Sun, 13 May 2012 14:54:00 GMT

摘要: 计算几何二分 + 求面�U?nbsp; 阅读全文

coreBugZJ 2012-05-13 22:54 发表评论

EOJ 1189 Wall POJ 1113 Wall

coreBugZJ — Sun, 13 May 2012 14:52:00 GMT

摘要: 计算几何凸包阅读全文

coreBugZJ 2012-05-13 22:52 发表评论

coreBugZJ — Sun, 13 May 2012 14:51:00 GMT

摘要: 多边形面�U�（计算几何�Q?nbsp; 阅读全文

coreBugZJ 2012-05-13 22:51 发表评论

POJ 3277 City Horizon

coreBugZJ — Sun, 22 Apr 2012 14:52:00 GMT

摘要: �U�段�?��L��化。。�?nbsp; 阅读全文

coreBugZJ 2012-04-22 22:52 发表评论

POJ 2528 Mayor's posters

coreBugZJ — Sun, 22 Apr 2012 14:50:00 GMT

摘要: �U�段�?��L��化。。�?nbsp; 阅读全文

coreBugZJ 2012-04-22 22:50 发表评论

EOJ 2458 Frequent values

coreBugZJ — Sun, 22 Apr 2012 14:48:00 GMT

摘要: �U�段树，�?RMQ ST 。。�?nbsp; 阅读全文

coreBugZJ 2012-04-22 22:48 发表评论

EOJ 2525 Light Switching

coreBugZJ — Sun, 22 Apr 2012 14:46:00 GMT

摘要: �U�段树。。�?nbsp; 阅读全文

coreBugZJ 2012-04-22 22:46 发表评论

coreBugZJ — Sat, 21 Apr 2012 08:59:00 GMT

摘要: 深度优先搜烦�Q�优化剪枝。。�?nbsp; 阅读全文

coreBugZJ 2012-04-21 16:59 发表评论

EOJ 1780 Escape

coreBugZJ — Sat, 21 Apr 2012 08:40:00 GMT

摘要: 搜烦题，本来是简单题�Q�但我犯了错误。。�?nbsp; 阅读全文

coreBugZJ 2012-04-21 16:40 发表评论

ACM搜烦题经�?Sticks

coreBugZJ — Sat, 21 Apr 2012 02:47:00 GMT

摘要: EOJ 1981 Sticks
POJ 1011 Sticks
HDOJ 1455 Sticks
UVA 307 Sticks 。。�?nbsp; 阅读全文

coreBugZJ 2012-04-21 10:47 发表评论

EOJ 2069 Asteroids

coreBugZJ — Fri, 30 Mar 2012 14:18:00 GMT

摘要: 求二分图最大匹配，使用匈牙利算法。。�?nbsp; 阅读全文

coreBugZJ 2012-03-30 22:18 发表评论

EOJ 1864 Playing With Cubes

coreBugZJ — Fri, 30 Mar 2012 14:16:00 GMT

摘要: 二分图最大匹配��用匈牙利��法。。�?nbsp; 阅读全文

coreBugZJ 2012-03-30 22:16 发表评论

EOJ 1096 ��盘分割 �Q�动态规划）

coreBugZJ — Sat, 17 Mar 2012 03:28:00 GMT

摘要: .
��一�?8*8 的棋盘进行如下分�Ԍ��
��原��盘割下一块矩形棋盘�ƈ使剩下部分也是矩形，再将剩下部分�l�箋如此分割�Q?
�q�样割了 n-1 �ơ后�Q�连同最后剩下的矩�Ş��盘共有 n 块矩形棋盘�?
每次切割都只能沿着��盘格子的边�q�行�?

原棋盘上每一格有一个分��|��一块矩形棋盘的��d��为其所含各格分��g��和�?
现需要把��盘按上�q�规则分割成 n 块矩形棋盘，�q��各矩形棋盘��d��的均方差最��?nbsp; 阅读全文

coreBugZJ 2012-03-17 11:28 发表评论

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微��Y2014实习生及�U���o营技术类职位在线���试

LZW �~�码解码代码

A* ���法求解八数码问题，POJ 1077 Eight

k-means ���法实现人口聚类

POJ 2068 Nim

POJ 2975 Nim

POJ 3696 The Luckiest number

POJ 3604 Professor Ben

EOJ 1117 剩余定理

POJ 1603 Risk

EOJ 1848 你是ACM吗？

EOJ 1708 Connected Gheeves

EOJ 1189 Wall POJ 1113 Wall

POJ 3277 City Horizon

POJ 2528 Mayor's posters

EOJ 2458 Frequent values

EOJ 2525 Light Switching

EOJ 1780 Escape

ACM搜烦题经�?Sticks

EOJ 2069 Asteroids

EOJ 1864 Playing With Cubes

EOJ 1096 ���盘分割 �Q�动态规划）

微��Y2014实习生及�U��o营技术类职位在线��试

A* ��法求解八数码问题，POJ 1077 Eight

k-means ��法实现人口聚类

EOJ 1096 ��盘分割 �Q�动态规划）