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Description: 函数int SingleTrackE(int i, int j, int iop, int *code) 实现单区域轮廓跟踪
函数int CodeToTable(int *code,struct DOT *dot)实现链码表转换成线段表 -function int SingleTrackE (int i, int j, iop int, int * code) to achieve single contour tracking function int CodeToTable (int * code, struct * dot DOT) achieve chain code table into segments Table
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Size: 2507 |
Author: 朱云 |
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Description: 图的遍历的演示（c 语言 数据结构课程设计题）
/*定义图*/
typedef struct{
int V[M]
int R[M][M]
int vexnum
}Graph
/*创建图*/
void creatgraph(Graph *g,int n)
{
int i,j,r1,r2
g->vexnum=n
/*顶点用i表示*/
for(i=1 i<=n i++)
{
g->V[i]=i
}
/*初始化R*/
for(i=1 i<=n i++)
for(j=1 j<=n j++)
{
g->R[i][j]=0
}
/*输入R*/
printf(\"Please input R(0,0 END):\\n\")
scanf(\"％d,％d\",&r1,&r2)
while(r1!=0&&r2!=0)
{
g->R[r1][r2]=1
g->R[r2][r1]=1
scanf(\"％d,％d\",&r1,&r2)
}
}
-graph traversal of the display (c language curriculum design data structures that) / * definition of the map * / typedef s truct V (int int [M] R [M] [M]) int vexnum Graph / * create map * / void creatgraph (Graph * g, int n) (int i, j, r1, r2 g -
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Size: 1405 |
Author: 吉庆 |
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Description: 数据结构（严慰敏）配套纯c代码实验十
typedef int InfoType // 定义其它数据项的类型
typedef int KeyType // 定义RedType类型的关键字为整型
struct RedType // 记录类型(同c10-1.h)
{
KeyType key // 关键字项
InfoType otherinfo // 其它数据项
}
typedef char KeysType // 定义关键字类型为字符型
#include\"c1.h\"
#include\"c10-3.h\"
void InitList(SLList &L,RedType D[],int n)
{ // 初始化静态链表L（把数组D中的数据存于L中）
char c[MAX_NUM_OF_KEY],c1[MAX_NUM_OF_KEY]
int i,j,max=D[0].key //
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Size: 13016 |
Author: 冰河 |
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Description: 五子棋小游戏#include<stdlib.h>
#include<time.h>
#include<stdio.h>
#include<conio.h>
int chess[169]={0} /*棋盘*/
struct chess_t/*作为辅助，即是作为建意*/
{
char attack /*攻防用的,0表示守,1表示攻*/
int j /*作为优先级用*/
}chess_a[169]
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Size: 2382 |
Author: 庄斌 |
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Description: 家 庭 财 务 管 理 小 程 序
程序源代码：
/*money management system*/
#include \"stdio.h\"
#include \"dos.h\"
main()
{
FILE *fp
struct date d
float sum,chm=0.0
int len,i,j=0
int c
Platform: |
Size: 1498 |
Author: earl86 |
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Description: 函数int SingleTrackE(int i, int j, int iop, int *code) 实现单区域轮廓跟踪
函数int CodeToTable(int *code,struct DOT *dot)实现链码表转换成线段表 -function int SingleTrackE (int i, int j, iop int, int* code) to achieve single contour tracking function int CodeToTable (int* code, struct* dot DOT) achieve chain code table into segments Table
Platform: |
Size: 2048 |
Author: 朱云 |
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Description: 图的遍历的演示（c 语言 数据结构课程设计题）
/*定义图*/
typedef struct{
int V[M]
int R[M][M]
int vexnum
}Graph
/*创建图*/
void creatgraph(Graph *g,int n)
{
int i,j,r1,r2
g->vexnum=n
/*顶点用i表示*/
for(i=1 i<=n i++)
{
g->V[i]=i
}
/*初始化R*/
for(i=1 i<=n i++)
for(j=1 j<=n j++)
{
g->R[i][j]=0
}
/*输入R*/
printf("Please input R(0,0 END):\n")
scanf("％d,％d",&r1,&r2)
while(r1!=0&&r2!=0)
{
g->R[r1][r2]=1
g->R[r2][r1]=1
scanf("％d,％d",&r1,&r2)
}
}
-graph traversal of the display (c language curriculum design data structures that)/* definition of the map*/typedef s truct V (int int [M] R [M] [M]) int vexnum Graph/* create map*/void creatgraph (Graph* g, int n) (int i, j, r1, r2 g-
Platform: |
Size: 1024 |
Author: 吉庆 |
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Description: 数据结构（严慰敏）配套纯c代码实验十
typedef int InfoType // 定义其它数据项的类型
typedef int KeyType // 定义RedType类型的关键字为整型
struct RedType // 记录类型(同c10-1.h)
{
KeyType key // 关键字项
InfoType otherinfo // 其它数据项
}
typedef char KeysType // 定义关键字类型为字符型
#include"c1.h"
#include"c10-3.h"
void InitList(SLList &L,RedType D[],int n)
{ // 初始化静态链表L（把数组D中的数据存于L中）
char c[MAX_NUM_OF_KEY],c1[MAX_NUM_OF_KEY]
int i,j,max=D[0].key // -Data structure (Yen Wei-Min) supporting pure c code experiment 10 typedef int InfoType// definition of other data types typedef int KeyType// definition of types of keywords for RedType integer struct RedType// record type (with c10-1.h) (KeyType key// keyword items InfoType otherinfo// other data item) typedef char KeysType// definition of a keyword type character# include c1.h# include c10-3.h void InitList (SLList
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Size: 13312 |
Author: 冰河 |
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Description: 五子棋小游戏#include<stdlib.h>
#include<time.h>
#include<stdio.h>
#include<conio.h>
int chess[169]={0} /*棋盘*/
struct chess_t/*作为辅助，即是作为建意*/
{
char attack /*攻防用的,0表示守,1表示攻*/
int j /*作为优先级用*/
}chess_a[169] -Gobang game# Include
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Size: 2048 |
Author: 庄斌 |
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Description: 家 庭 财 务 管 理 小 程 序
程序源代码：
/*money management system*/
#include "stdio.h"
#include "dos.h"
main()
{
FILE *fp
struct date d
float sum,chm=0.0
int len,i,j=0
int c
-Family financial management applet source code:/* money management system* /# Include stdio.h# Include dos.h main () (FILE* fp struct date d float sum, chm = 0.0 int len, i, j = 0 int c
Platform: |
Size: 1024 |
Author: earl86 |
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Description: #include <stdlib.h>
#include <stdio.h>
#include <time.h>
#include <fstream>
#include <iostream>
using namespace std
struct Code
{
float a
char s
}
Code cc[8]={{0.25, a },{0.3, b },{0.1, c },{0.05, d },{0.05, e },{0.05, . },{0.05, , },{0.15, 8 }}
int main()
{
srand((unsigned)time(NULL))
int x
int n
ofstream outobj("plpl.txt")
cout<<"inpute the number(要产生的字符数):"
cin>>n
for(int i=0 i<n i++)
{
x=rand()％1000+1
int low =1
for(int j=0 j<8 j++)
{
int high=cc[j].a*1000+low-1
if(x>=low&&x<=high)
{
outobj<<cc[j].s
break
}
low = high+1
}
}
outobj.flush()
outobj.close()
} 存到.dat文件里面-# Include <stdlib.h># Include <stdio.h># Include <time.h># Include <fstream># Include <iostream> using namespace std struct Code (float a char s) Code cc [ 8] = ((0.25, a), (0.3, b), (0.1, c), (0.05, d), (0.05, e), (0.05,.), (0.05,,), (0.15, 8 )) int main () (srand ((unsigned) time (NULL)) int x int n ofstream outobj ( plpl.txt ) cout << inpute the number (to generate the number of characters): cin>> n for (int i = 0 i <n i++) (x = rand () 1000+ 1 int low = 1 for (int j = 0 j <8 j++) (int high = cc [j] . a* 1000+ low-1 if (x> = low
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Size: 1024 |
Author: 糊涂 |
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Description: spring struct2 hibernate 整合的一个例子，有完整jar包-spring struct2 hibernate demo
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Size: 18991104 |
Author: prl |
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Description: extjs MVC struct 开发框架，快速开发应用-development framework extjs MVC struct
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Size: 158720 |
Author: chen |
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Description: 本文的目的，是希望通过跟各位朋友一起讨论关于指针的几个关键概念及常见问题，以加深对指针的理解。因此，本文并不是讲述形如int *p、struct {int i float j } *p等这些东西是什么的文章，阅读本文的朋友最好对指针已经具有一定的使用经验，正因如此，笔者才给文章起名叫《再再论指针》。-This book is written to discuss something we may don t know about the pointer in the C programming language
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Size: 29696 |
Author: luciola |
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Description: 图的遍历的演示（c 语言 数据结构课程设计题） /*定义图*/ typedef struct{ int V[M] int R[M][M] int vexnum }Graph /*创建图*/ void creatgraph(Graph *g,int n) { int i,j,r1,r2 g->vexnum=n /*顶点用i表示*/ for(i=1 i<=n i++) { g->V[i]=i } /*初始化R*/ for(i=1 i<=n i++) for(j=1 j<=n j++) { g->R[i][j]=0 } /*输入R*/ printf("Please input R(0,0 END):\n") scanf("％d,％d",&r1,&r2) while(r1!=0&&r2!=0) { g->R[r1][r2]=1 g->R[r2][r1]=1 scanf("％d,％d",&r1,&r2) } } -graph traversal of the display (c language curriculum design data structures that)/* definition of the map*/typedef s truct V (int int [M] R [M] [M]) int vexnum Graph/* create map*/void creatgraph (Graph* g, int n) (int i, j, r1, r2 g--Graph traversal of the presentation (c language data structure, curriculum design issues) /* definition of map*/typedef struct (int V [M] int R [M] [M] int vexnum) Graph /* create the map*/void creatgraph (Graph* g, int n) (int i, j, r1, r2 g-> vexnum = n /* vertex with i said*/for (i = 1 i < = n i++) (g-> V [i] = i ) /* initialize the R*/for (i = 1 i < = n i++) for (j = 1 j < = n j++) (g-> R [i] [j] = 0) /* input R*/printf (" Please input R (0,0 END): \ n" ) scanf (" d, d" , & r1, & r2) while (r1! = 0 & & r2! = 0) (g-> R [r1] [r2] = 1 g-> R [r2] [r1] = 1 scanf (" d, d" , & r1, & r2)))-graph traversal of the display (c language curriculum design data structures that) /* definition of the map*/typedef s truct V (int int [M] R [M] [M]) int vexnum Graph /* create map*/void creatgraph (Graph* g, int n) (int i, j, r1 , r2 g-
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Size: 60416 |
Author: 唐钊 |
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Description: typedef struct OLNode
{
int i,j
ElemType e
struct OLNode *right,*down
} OLNode, *OLink
typedef struct
{
OLink *rhead,*chead
int mu,nu,tu
} CrossList
十字链表法，求解稀疏矩阵运算，程序容易看懂。不过没有乘法-typedef struct OLNode
{
int i,j
ElemType e
struct OLNode*right,*down
} OLNode,*OLink
typedef struct
{
OLink*rhead,*chead
int mu,nu,tu
} CrossList
Platform: |
Size: 7168 |
Author: zzc |
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Description: NStepSCAN N步磁盘扫描调度算法C++程序-#include "stdafx.h"
#include "stdlib.h"
int L,P
int a[20]
int flag //访问位
typedef struct TASK
{
int requrie
int io
} TASK
TASK T[50][50]
TASK N[50]
int diskway //当前磁道
void iodowith()
void init()
{
int i,j
for(i=0 i<20 i++)
{
for(j=0 j<20 j++)
{
T[i][j].requrie=-1
}
}
for(i=0 i<L i++)
{
for(j=0 j<P j++)
{
T[i][j].requrie=rand() 200
T[i][j].io=rand() 2
}
}
}
void output()
{
printf("磁盘访问的初始顺序是：\n")
for(int i=0 i<L i++)
{
printf("第 d组是:\n",i+1)
for(int j=0 j<P j++)
{
printf(" d d\t",T[i][j].requrie,T[i][j].io)
}
printf("\n")
}
}
void sort_btos(TASK x[],int n)
{
int i,j,k
TASK t
for(i=0 i<n-1 i++)
{
k=i
for(j=i+1 j<n j++)
if(x[j].requrie>x[k].requrie)
k=j
if(k!=i)
{
t=x[i]
x[i]=x[k]
x[k]=t
}
}
}
void sort_stob(TASK x[],int n)
{ int i,j,k
TASK t
fo
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Size: 1024 |
Author: 吴英杰 |
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Description: 用c++编写的选票系统，并可统计相关得票情况-#include <iostream>
using namespace std
struct Person//声明结构体类型Person；
{
char name[20]
int count
}
int main()
{
Person leader[3]={"aaa",0,"bbb",0,"ccc",0}
//定义Person类型的数组，内容为当前候选人的姓名及得票数；
char leader_name[20] //leader_name为投票人所选的人姓名；
int i,j
for(i=0 i<10 i++)
{
cin>>leader_name //先后输入十张票上所选的人的姓名；
for(j=0 j<3 j++)
{
if(strcmp(leader_name,leader[j].name)==0)
leader[j].count++ //所选人与候选人的姓名相同，则该候选人的票数加1；
}
}
cout<<endl
double round
for(i=0 i<3 i++)
{
round=leader[i].count/10
cout<<leader[i].name<<":"<<"得票数:"<<leader[i].count<<"得票率:"<<round<<endl
}
return 0
}
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Size: 854016 |
Author: 盛雯雯 |
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Description: 线性表链式存储（双向链表）插入、删除运算
1、预习要求：线性表的插入、删除相关概念及运算，完成线性表元素的插入、删除。
2、实验目的：
（1）了解线性表的插入、删除相关概念；
（2）理解线性表的插入、删除过程和结构定义；
（3）掌握算法转换为程序的过程中的变化。
3、实验内容及要求：
(1) 以“一个较完整的程序范例-顺序表”程序框架为框架
(2）双链表结构体为：
typedef struct
{
int num /学号
char name[10] /姓名
} student；
分别建立包含3~5个数据元素的链式存储线性表；
(3）实现功能
a)构造空表，直接从键盘输入数据元素
b) 插入、删除、一般查找（给定某一学号x，找到并显示该学生信息）、显示等函数功能
c）设计特殊查找函数findbeforeAndAfter(...)，给定某一学号x，能同时找到该学生第i个前驱的学生信息，以及该学生第j个后继的学生信息，参数x, i，j从键盘输入
（4）给出程序运行截图。
-The linear table chain store (doubly linked list) insert, delete operations
1, the preview request: Linear table insert, delete the related concepts and operations, completion of the linear table elements
Insert, delete.
2, the purpose of the experiment:
(1) to understand the linear table insert, delete the related concepts
(2) understanding of the linear table insert, delete, processes and structures defined
(3) to grasp the changes in the process of algorithm into a program.
3, experimental content and requirements:
(1) "a more complete example of the program- order form" procedural framework within the framework of
(2) doubly linked list structure is:
typedef struct
{
int num /Student ID
char name [10] /Name
} Student
Were established to contain three to five data elements in the chain store linear table
(3) to achieve functional
a) Construct an empty table directly from the keyboard to input data elements
b)
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Size: 325632 |
Author: the spring |
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Description: 与矩阵连乘最小乘法数问题类似，虽然不需要记录加括号的方式，但是需要纪录下从i乘到j的结果中为a、b、c的个数，以便在后续的计算中可以用到之前的数据，避免重复计算。所以就需要用三维数组来存储（当然用struct结构体数据类型也很方便）或者用两个数组。-With a matrix with minimal multiplication problem similar, although need not record bracketed style, but need to record from I to take the j results for a, B, C numbers, in order to follow-up calculation can be used before the data, avoid to repeat computation. So we need to use three-dimensional array to store ( of course with the structure of the struct data type is also very convenient ) or a two array.
Platform: |
Size: 295936 |
Author: 张小迪 |
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