4数据结构上机考试,在已有的数据库中修改数据

作者: 新金沙平台  发布:2019-09-22

  1. 透过顾客输入建设构造一棵树,程序应能正确输出该树的节点数,以 验证输入的树是或不是科学;


Unix & Linux历史源流

开始时代Linux的开机管理程序(boot loader)使用LILO(Linux Loader),开始的一段年代的LILO存在着有个别难以容忍的缺点,举例不可能识别1024柱面未来的硬盘空间,后来的GRUB(GRand Unified Bootloader)制服这几个老毛病,具备‘动态搜索内核文件’的效果与利益,能够让客商在开机的时候,自行编排开机设置系统文件,通过ext2或ext3文件系统中加载Linux Kernel(GRUB通过差别的文件系统驱动能够分辨差相当少具有Linux协理的文件系统,因而得以选拔过多文件系统来格式化内核文件所在的扇区,并不囿于于ext文件系统)。
Linux的证明和吉祥物是叁只名字叫做Tux的企鹅,标识的原因是因为Linus在澳大南宁(Australia)时曾被多头动物园里的企鹅咬了一口,便选用企鹅作为Linux的阐明。更便于被接受的布道是:企鹅代表南极,而南极又是满世界所共有的一块陆地。那也就代表Linux是所有人的Linux。

测量试验用的树

UNIT_PEnclaveICE :数据库原先类型为:NUMBE福睿斯

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--3:删除原本的字段
ALTER TABLE CLOUD_RELEASE.CONTRACT_PRODUCT_ITEM DROP COLUMN UNIT_PRICE;

BSD

386BSD因为法律难题直到一九九四年还尚未发布,NetBSD和FreeBSD是386BSD的后生,早于Linux。林纳斯·托瓦兹曾说,当时只要有可用的386BSD,他就也许不会编写Linux。

  2.函数模块化,使函数尽量轻巧易读

急需:将以上的三个字段的数据类型保留位数修改为6位,该表中已有数量,并且那七个*字段中也会有值(原先的数量须求保留),方法如下:**

GNU

Richard·马特hew·Stowe曼(Richard M. Stallman),GNU陈设的创小编。
1984年,Richard·马特hew·Stowe曼创建GNU陈设。这一个安插有多个指标,是为着升高中二年级个截然自由的类Unix操作系统。自1983年倡导那么些布署以来,在一九八三年,理查德·马特hew·Stowe曼发起自由软件基金会并且在1990年创作GPL。一九九〇年间早期,GNU开始多量的发生或搜罗各类系统所须要的零件,疑似——库、编写翻译器、调节和测量检验工具、文本编辑器、网页服务器,以及一个Unix的客商分界面(Unix shell)——但是像某些底层情况,如硬件驱动、守护进度运营基础还是不完整和深陷停顿,GNU安排中是在马赫先生微核(Mach microkernel)的架构之上开采种类基本,也便是所谓的GNU Hurd,可是这几个基于Mach的统一图谋非常复杂,发展进程则相对缓慢。林纳斯·托瓦兹曾说过假设GNU内核在1995年时得以用,他不会友善去写三个。

  1 #include<iostream>  2 #include<string>  3 #include<stack>  4 #include<queue>  5 using namespace std;  6   7 class FTree {  8 public:  9     FTree(); 10     void getNodeNum(); 11     //递归先序构造二叉树形式的树/森林 12     FTree(string str, int& index); 13     //2)树和森林的先根遍历的递归和迭代算法; 14     void preOrderbyRecursive(); 15     void preOrderbyIteration(); 16     //3)树和森林的后根遍历的递归和迭代算法; 17     void afterOrderbyRecursive(); 18     void afterOrderbyIteration(); 19     //4)树和森林的层次遍历算法。 20     void levelOrder(); 21     void getReltion(char node1,char node2); 22     void getReltion(); 23     FTree* getNode(char node); 24     ~FTree(); 25     bool isFatherOf(char node); 26     bool isBrotherOf(char node); 27     bool isGrandFatherOf(char node); 28     bool isOtherTree(char node1,char node2); 29     bool isSameTree(char node); 30     bool haveLittleSon(char node); 31 protected: 32     char data; 33     FTree* left_son; 34     FTree* right_brother; 35     int node_num; 36 }; 37  38 int main() { 39  40     //string ss = "124##57###3#6##"; int index = 0; 41     //string ss = "RAB#C#D##EF##GH#IJ#####";  42     string ss; 43     cin >> ss; 44     int index = 0; 45     FTree t(ss, index); 46     t.getNodeNum(); 47     cout << "从字符串 " << ss << " 构造树t " << endl; 48     //t.getReltion; 49     while (1) { 50         t.getReltion(); 51     } 52  53 } 54  55  56  57 FTree::FTree() 58 { 59     data = NULL; 60     left_son = 0; 61     right_brother = 0; 62 } 63 void FTree::getNodeNum() 64 { 65     cout << "树的节点个数为" << node_num << endl; 66 } 67 FTree::FTree(string str, int& index) 68 { 69     if (str[index] == NULL) { 70         cout << "输入的字符串错误!"; 71     } 72     node_num = 0; 73     if (index <= str.size() - 1 && str[index] != '#') { 74         data = str[index]; 75         index  ; 76         node_num  ; 77         if (index <= str.size() - 1) { 78             if (str[index] == '#') { 79                 left_son = 0; 80                 index  ; 81             } 82             else { 83                 left_son = new FTree(str, index); 84                 node_num  = left_son->node_num; 85             } 86         } 87         if (index <= str.size() - 1) { 88             if (str[index] == '#') { 89                 right_brother = 0; 90                 index  ; 91             } 92             else { 93                 right_brother = new FTree(str, index); 94                 node_num  = right_brother->node_num; 95             } 96         } 97     } 98     else { 99         data = NULL;100         index  ;101         left_son = 0;102         right_brother = 0;103     }104 }105 106 void FTree::preOrderbyRecursive()107 {108     if (data == NULL) {109         return;110     }111     cout << data;112     if (left_son != 0) {113         left_son->preOrderbyRecursive();114     }115     if (right_brother != 0) {116         right_brother->preOrderbyRecursive();117     }118 }119 120 void FTree::preOrderbyIteration()121 {122     stack<FTree*> sta;123     if (data == NULL) { return; }124     cout << data;125     if (left_son != 0) { sta.push; }126     if (right_brother != 0) { sta.push(right_brother); }127     while (!sta.empty {128         FTree* f = sta.top(); sta.pop();129         cout << f->data;130         if (right_brother != 0) { sta.push(right_brother); }131         if (left_son != 0) { sta.push; }132     }133 }134 135 void FTree::afterOrderbyRecursive()136 {137     if (data == NULL) {138         return;139     }140 141     if (left_son != 0) {142         left_son->afterOrderbyRecursive();143     }144     cout << data;145     if (right_brother != 0) {146         right_brother->afterOrderbyRecursive();147     }148 149 }150 151 void FTree::afterOrderbyIteration()152 {153     stack<FTree*> sta;154     if (data == NULL) { return; }155     if (right_brother != 0) { sta.push(right_brother); }156     sta.push(this);157     if (left_son != 0) { sta.push; }158     while (!sta.empty {159         FTree* f = sta.top(); sta.pop();160         if (right_brother != 0) { sta.push(right_brother); }161         cout << f->data;162         if (left_son != 0) { sta.push; }163     }164 }165 166 void FTree::levelOrder()167 {168     queue<FTree*> que;169     que.push(this);170     while (!que.empty {171         FTree* f = que.front();    que.pop();172         while (f != 0) {173             cout << f->data;174             if (f->left_son != 0) {175                 que.push(f->left_son);176             }177             f = f->right_brother;178         }179 180     }181 }182 183 void FTree::getReltion(char node1, char node2)184 {185     FTree* no1 = getNode;186     FTree* no2 = getNode;187     if (no1->isFatherOf == 1) {188         return;189     }190     if (no2->isFatherOf == 1) {191         return;192     }193     if (no1->isGrandFatherOf == 1) {194         return;195     }196     if (no2->isGrandFatherOf == 1) {197         return;198     }199 200     if (no1->isBrotherOf == 1) {201         return;202     }203     if (no2->isBrotherOf == 1) {204         return;205     }206     if (this->isOtherTree(node1,node2) == 1) {207         return;208     }209     if (no1->isSameTree == 1) {210         return;211     }212     if (no2->isSameTree == 1) {213         return;214     }215     cout << "其他关系" << endl;216 }217 218 void FTree::getReltion()219 {220     char node1, node2;221     cout << "nplease input node1,node2n";222     cin >> node1;223     cin>>node2;224     getReltion(node1, node2);225 }226 227 FTree * FTree::getNode(char node)228 {229     if (data == node) {230         return this;231     }232     FTree* find = 0;233     FTree* current = left_son;234     while (current != 0) {235         find = current->getNode;236         if (find != nullptr) {237             return find;238         }239         current = current->right_brother;240     }241     return 0;242     return find;243 }244 245 FTree::~FTree()246 {247     if (left_son != 0) {248         //cout << left_son->data;249         delete left_son;250     }251     if (right_brother != 0) {252         //cout << right_brother->data;253         delete right_brother;254     }255 256 }257 258 bool FTree::isFatherOf(char node)259 {260     261     FTree* current = left_son;262     while (current != 0) {263         if (current->data == node) {264             cout << data << "与" << node << "是父子关系,";265             cout << data << "是" << node << "的父亲"<<endl;266             return 1;267         }268         else {269             current = current->right_brother;270         }271     }272     return false;273 }274 275 bool FTree::isBrotherOf(char node)276 {277     FTree* current = right_brother;278     while (current != 0) {279         if (current->data == node) {280             cout << data << "与" << node << "是兄弟关系n";281             cout << data << "是" << node << "的兄长";282             return 1;283         }284         else {285             current = current->right_brother;286         }287     }288     return false;289 }290 291 bool FTree::isGrandFatherOf(char node)292 {293     FTree* current = left_son;294     while (current != 0) {295         if (current->left_son != 0) {296             FTree* currentson = current->left_son;297             while (currentson != 0) {298                 if (currentson->data == node) {299                     cout << data << "与" << node << "是祖孙关系n";300                     cout << data << "是" << node << "的祖父";301                     return 1;302                 }303                 currentson = currentson->right_brother;304             }305         }306         current = current->right_brother;307     }308     return false;309 }310 311 bool FTree::isOtherTree(char node1, char node2)312 {313     int flag = 0;314     FTree* current = left_son;315     while (current != 0) {316         if (current->haveLittleSon{317             flag  ;318         }319         else {320             if (current->haveLittleSon {321                 flag  ;322             }323         }324         current = current->right_brother;325     }326     if (flag == 2) {327         cout << "其他关系,"<< node1 << "与" << node2<<"不在同一分支上,"   <<endl;328         return 1;329     }330     return false;331 }332 333 bool FTree::isSameTree(char node)334 {335 336     FTree* current = left_son;337     while (current != 0) {338         if (current->left_son != 0) {339             if (current->haveLittleSon {340                 cout << "其他关系,在同一分支上," << data << "是" << node << "祖父以上长辈"<<endl;341                 return 1;342             }343         }344         current = current->right_brother;345     }346     return false;347 }348 349 bool FTree::haveLittleSon(char node)350 {351     if (data == node) {352         return 1;353     }354     bool flag=0;355     FTree* current = left_son;356     while (current != 0) {357         if (current->data == node) {358             return 1;359         }360         flag = current->haveLittleSon;361         current = current->right_brother;362     }363     return flag;364 }

AMOUNT :数据库原先类型为:NUMBE本田CR-V

写在终极的话

Linux代表着分享精神,本着这种精神,小编创立了Linux高校网(www.linuxdaxue.com),希望能扶助到那二个想要学习Linux的爱人。

  对于左外孙子—右兄弟链接存储的树,任给树中七个节点,判别双方之间的涉及属于祖孙、老爹和儿子、兄弟、其余关系中的哪一种关系。注意,叁个节点是别的贰个节点的幼子节点的幼子节点则称那多个节点之间为 祖孙关系。设计的程序应变成如下的意义:

ALTER TABLE CLOUD_RELEASE.CONTRACT_PRODUCT_ITEM MODIFY(UNIT_PRICE NUMBER;

Linux历史

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