Insurance A Vehicle Vehicle Insured In Accident Vehicle Inspected Online fprms goto insurer Claim Approved Vehicle Repaired Insurer Pays All data should be saved to the data Base
标签： Vehicle Insurance Inspected Accident
上传时间： 2017-09-01
上传用户：xz85592677
Parallel robotic manipulators can be considered a well-established option for many different applications of manipulation, machining, guiding, testing, control, tracking, haptic force feed-back, etc. A typical parallel robotic manipulator (PM) consists of a mobile platform connected to the Base (fixed platform) by at least two kinematic chains called limbs. The mobile platform can achieve between one and three independent translations (T) and one to three independent rotations (R).
标签： well-established manipulators considered different
上传时间： 2017-09-03
上传用户：moerwang
Per gli interessati ai metodi della compressione una vera miniera d oro, oltre 70 algoritmi all interno di moduli Base indipendenti consentono a questo programma di mostrare il loro utilizzo e le loro performances, ecco elencati alcuni di essi : Base64/crc32/fibonacci/mtf/freq/adddif/bwt/fix12/fix128/flatter/ huffman/lzw/lzs/rle/lbe/hash/vbc/scrambler, e tanti tanti altri.
标签： compressione interessati algoritmi miniera
上传时间： 2013-12-16
上传用户：时代电子小智
vc++与visio 安装VISIO office（推荐2003以上版本）必须安装visio office 否则程序无法运行 安装VisioSDK（推荐2003以上版本） 默认安装路径为<SDK> C:\Program Files\Microsoft Office\Office12\VisSDK 新建project（实例代码是新建了个MFC Dialog Base program） 将<SDK>\Libraries\CPP\里的include目录下和sources目录下的如下文件拷贝到你的project目录里（这样可以。。）
标签： office visio 2003 VisioSDK
上传时间： 2013-12-25
上传用户：大融融rr
as a message came into prominence with the publication in 1948 of an influential paper by Claude Shannon, "A Mathematical Theory of Communication." This paper provides the foundations of information theory and endows the word information not only with a technical meaning but also a measure. If the sending device is equally likely to send any one of a set of N messages, then the preferred measure of "the information produced when one message is chosen from the set" is the Base two logarithm of N (This measure is called self-information). In this paper, Shannon cont
标签： influential publication prominence message
上传时间： 2014-01-21
上传用户：2404
My JSP 'TeacherMain.jsp' starting page var $=function(id) { return document.getElementById(id); } function show_menu(num){ for(i=0;i
标签： C++
上传时间： 2015-07-03
上传用户：xiyuzhu
#include <stdlib.h> #include<stdio.h> #include <malloc.h> #define stack_init_size 100 #define stackincrement 10 typedef struct sqstack { int *Base; int *top; int stacksize; } sqstack; int StackInit(sqstack *s) { s->Base=(int *)malloc(stack_init_size *sizeof(int)); if(!s->Base) return 0; s->top=s->Base; s->stacksize=stack_init_size; return 1; } int Push(sqstack *s,int e) { if(s->top-s->Base>=s->stacksize) { s->Base=(int *)realloc(s->Base,(s->stacksize+stackincrement)*sizeof(int)); if(!s->Base) return 0; s->top=s->Base+s->stacksize; s->stacksize+=stackincrement; } *(s->top++)=e; return e; } int Pop(sqstack *s,int e) { if(s->top==s->Base) return 0; e=*--s->top; return e; } int stackempty(sqstack *s) { if(s->top==s->Base) { return 1; } else { return 0; } } int conversion(sqstack *s) { int n,e=0,flag=0; printf("输入要转化的十进制数：\n"); scanf("%d",&n); printf("要转化为多少进制：\n"); scanf("%d",&flag); printf("将十进制数%d 转化为%d 进制是：\n",n,flag); while(n) { Push(s,n%flag); n=n/flag; } while(!stackempty(s)) { e=Pop(s,e); switch(e) { case 10: printf("A"); break; case 11: printf("B"); break; case 12: printf("C"); break; case 13: printf("D"); break; case 14: printf("E"); break; case 15: printf("F"); break; default: printf("%d",e); } } printf("\n"); return 0; } int main() { sqstack s; StackInit(&s); conversion(&s); return 0; }
上传时间： 2016-12-08
上传用户：爱你198
满足混合字符串（汉字和数字等字符）批量（非一个字符一个字符），转换为16进制；同样支持16进制转换为字符串，C++代码； 在VS2010上用MFC编码测试可运行。可用于串口通信数据编码。
上传时间： 2017-05-31
上传用户：西蒙贝克
#include <iostream> #include <stdio.head> #include <stdlib.head> #include <string.head> #define ElemType int #define max 100 using namespace std; typedef struct node1 { ElemType data; struct node1 *next; }Node1,*LinkList;//链栈 typedef struct { ElemType *Base; int top; }SqStack;//顺序栈 typedef struct node2 { ElemType data; struct node2 *next; }Node2,*LinkQueue; typedef struct node22 { LinkQueue front; LinkQueue rear; }*LinkList;//链队列 typedef struct { ElemType *Base; int front,rear; }SqQueue;//顺序队列 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 //1．采用链式存储实现栈的初始化、入栈、出栈操作。 LinkList CreateStack()//创建栈 { LinkList top; top=NULL; return top; } bool StackEmpty(LinkList s)//判断栈是否为空，0代表空 { if(s==NULL) return 0; else return 1; } LinkList Pushead(LinkList s,int x)//入栈 { LinkList q,top=s; q=(LinkList)malloc(sizeof(Node1)); q->data=x; q->next=top; top=q; return top; } LinkList Pop(LinkList s,int &e)//出栈 { if(!StackEmpty(s)) { printf("栈为空。"); } else { e=s->data; LinkList p=s; s=s->next; free(p); } return s; } void DisplayStack(LinkList s)//遍历输出栈中元素 { if(!StackEmpty(s)) printf("栈为空。"); else { wheadile(s!=NULL) { cout<<s->data<<" "; s=s->next; } cout<<endl; } } 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 //2．采用顺序存储实现栈的初始化、入栈、出栈操作。 int StackEmpty(int t)//判断栈S是否为空 { SqStack.top=t; if (SqStack.top==0) return 0; else return 1; } int InitStack() { SqStack.top=0; return SqStack.top; } int pushead(int t,int e) { SqStack.top=t; SqStack.Base[++SqStack.top]=e; return SqStack.top; } int pop(int t,int *e)//出栈 { SqStack.top=t; if(!StackEmpty(SqStack.top)) { printf("栈为空."); return SqStack.top; } *e=SqStack.Base[s.top]; SqStack.top--; return SqStack.top; } 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 //3．采用链式存储实现队列的初始化、入队、出队操作。 LinkList InitQueue()//创建 { LinkList head; head->rear=(LinkQueue)malloc(sizeof(Node)); head->front=head->rear; head->front->next=NULL; return head; } void deleteEle(LinkList head,int &e)//出队 { LinkQueue p; p=head->front->next; e=p->data; head->front->next=p->next; if(head->rear==p) head->rear=head->front; free(p); } void EnQueue(LinkList head,int e)//入队 { LinkQueue p=(LinkQueue)malloc(sizeof(Node)); p->data=e; p->next=NULL; head->rear->next=p; head->rear=p; } 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 //4．采用顺序存储实现循环队列的初始化、入队、出队操作。 bool InitQueue(SqQueue &head)//创建队列 { head.data=(int *)malloc(sizeof(int)); head.front=head.rear=0; return 1; } bool EnQueue(SqQueue &head,int e)//入队 { if((head.rear+1)%MAXQSIZE==head.front) { printf("队列已满\n"); return 0; } head.data[head.rear]=e; head.rear=(head.rear+1)%MAXQSIZE; return 1; } int QueueLengthead(SqQueue &head)//返回队列长度 { return (head.rear-head.front+MAXQSIZE)%MAXQSIZE; } bool deleteEle(SqQueue &head,int &e)//出队 { if(head.front==head.rear) { cout<<"队列为空！"<<endl; return 0; } e=head.data[head.front]; head.front=(head.front+1)%MAXQSIZE; return 1; } int gethead(SqQueue head)//得到队列头元素 { return head.data[head.front]; } int QueueEmpty(SqQueue head)//判断队列是否为空 { if (head.front==head.rear) return 1; else return 0; } void travelQueue(SqQueue head)//遍历输出 { wheadile(head.front!=head.rear) { printf("%d ",head.data[head.front]); head.front=(head.front+1)%MAXQSIZE; } cout<<endl; } 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 //5．在主函数中设计一个简单的菜单，分别测试上述算法。 int main() { LinkList top=CreateStack(); int x; wheadile(scanf("%d",&x)!=-1) { top=Pushead(top,x); } int e; wheadile(StackEmpty(top)) { top=Pop(top,e); printf("%d ",e); }//以上是链栈的测试 int top=InitStack(); int x; wheadile(cin>>x) top=pushead(top,x); int e; wheadile(StackEmpty(top)) { top=pop(top,&e); printf("%d ",e); }//以上是顺序栈的测试 LinkList Q; Q=InitQueue(); int x; wheadile(scanf("%d",&x)!=-1) { EnQueue(Q,x); } int e; wheadile(Q) { deleteEle(Q,e); printf("%d ",e); }//以上是链队列的测试 SqQueue Q1; InitQueue(Q1); int x; wheadile(scanf("%d",&x)!=-1) { EnQueue(Q1,x); } int e; wheadile(QueueEmpty(Q1)) { deleteEle(Q1,e); printf("%d ",e); } return 0; }
上传时间： 2018-05-09
上传用户：123456..
obot control, a subject aimed at making robots behave as desired, has been extensively developed for more than two decades. Among many books being published on this subject, a common feature is to treat a robot as a single system that is to be controlled by a variety of control algorithms depending on different scenarios and control objectives. However, when a robot becomes more complex and its degrees of freedom of motion increase substantially, the needed control computation can easily go beyond the scope a modern computer can handle within a pre-specified sampling period. A solution is to Base the control on subsystem dynamics.
标签： decomposition virtual control
上传时间： 2019-09-04
上传用户：txb96
