<nav id="gkycu"></nav> 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
標簽： Vehicle Insurance Inspected Accident
上傳時間： 2017-09-01
上傳用戶：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).
標簽： well-established manipulators considered different
上傳時間： 2017-09-03
上傳用戶：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.
標簽： compressione interessati algoritmi miniera
上傳時間： 2013-12-16
上傳用戶：時代電子小智
vc++與visio 安裝VISIO office（推薦2003以上版本）必須安裝visio office 否則程序無法運行 安裝VisioSDK（推薦2003以上版本） 默認安裝路徑為<SDK> C:\Program Files\Microsoft Office\Office12\VisSDK 新建project（實例代碼是新建了個MFC Dialog BASE program） 將<SDK>\Libraries\CPP\里的include目錄下和sources目錄下的如下文件拷貝到你的project目錄里（這樣可以。。）
標簽： office visio 2003 VisioSDK
上傳時間： 2013-12-25
上傳用戶：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
標簽： influential publication prominence message
上傳時間： 2014-01-21
上傳用戶：2404
My JSP 'TeacherMain.jsp' starting page var $=function(id) { return document.getElementById(id); } function show_menu(num){ for(i=0;i
標簽： C++
上傳時間： 2015-07-03
上傳用戶：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("輸入要轉化的十進制數：\n"); scanf("%d",&n); printf("要轉化為多少進制：\n"); scanf("%d",&flag); printf("將十進制數%d 轉化為%d 進制是：\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; }
上傳時間： 2016-12-08
上傳用戶：愛你198
滿足混合字符串（漢字和數字等字符）批量（非一個字符一個字符），轉換為16進制；同樣支持16進制轉換為字符串，C++代碼； 在VS2010上用MFC編碼測試可運行。可用于串口通信數據編碼。
上傳時間： 2017-05-31
上傳用戶：西蒙貝克
#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;//鏈棧 typedef struct { ElemType *BASE; int top; }SqStack;//順序棧 typedef struct node2 { ElemType data; struct node2 *next; }Node2,*LinkQueue; typedef struct node22 { LinkQueue front; LinkQueue rear; }*LinkList;//鏈隊列 typedef struct { ElemType *BASE; int front,rear; }SqQueue;//順序隊列 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．采用鏈式存儲實現棧的初始化、入棧、出棧操作。 LinkList CreateStack()//創建棧 { LinkList top; top=NULL; return top; } bool StackEmpty(LinkList s)//判斷棧是否為空，0代表空 { if(s==NULL) return 0; else return 1; } LinkList Pushead(LinkList s,int x)//入棧 { 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)//出棧 { if(!StackEmpty(s)) { printf("棧為空。"); } else { e=s->data; LinkList p=s; s=s->next; free(p); } return s; } void DisplayStack(LinkList s)//遍歷輸出棧中元素 { if(!StackEmpty(s)) printf("棧為空。"); 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．采用順序存儲實現棧的初始化、入棧、出棧操作。 int StackEmpty(int t)//判斷棧S是否為空 { 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)//出棧 { SqStack.top=t; if(!StackEmpty(SqStack.top)) { printf("棧為空."); 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．采用鏈式存儲實現隊列的初始化、入隊、出隊操作。 LinkList InitQueue()//創建 { LinkList head; head->rear=(LinkQueue)malloc(sizeof(Node)); head->front=head->rear; head->front->next=NULL; return head; } void deleteEle(LinkList head,int &e)//出隊 { 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)//入隊 { 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．采用順序存儲實現循環隊列的初始化、入隊、出隊操作。 bool InitQueue(SqQueue &head)//創建隊列 { head.data=(int *)malloc(sizeof(int)); head.front=head.rear=0; return 1; } bool EnQueue(SqQueue &head,int e)//入隊 { if((head.rear+1)%MAXQSIZE==head.front) { printf("隊列已滿\n"); return 0; } head.data[head.rear]=e; head.rear=(head.rear+1)%MAXQSIZE; return 1; } int QueueLengthead(SqQueue &head)//返回隊列長度 { return (head.rear-head.front+MAXQSIZE)%MAXQSIZE; } bool deleteEle(SqQueue &head,int &e)//出隊 { if(head.front==head.rear) { cout<<"隊列為空！"<<endl; return 0; } e=head.data[head.front]; head.front=(head.front+1)%MAXQSIZE; return 1; } int gethead(SqQueue head)//得到隊列頭元素 { return head.data[head.front]; } int QueueEmpty(SqQueue head)//判斷隊列是否為空 { if (head.front==head.rear) return 1; else return 0; } void travelQueue(SqQueue head)//遍歷輸出 { 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．在主函數中設計一個簡單的菜單，分別測試上述算法。 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); }//以上是鏈棧的測試 int top=InitStack(); int x; wheadile(cin>>x) top=pushead(top,x); int e; wheadile(StackEmpty(top)) { top=pop(top,&e); printf("%d ",e); }//以上是順序棧的測試 LinkList Q; Q=InitQueue(); int x; wheadile(scanf("%d",&x)!=-1) { EnQueue(Q,x); } int e; wheadile(Q) { deleteEle(Q,e); printf("%d ",e); }//以上是鏈隊列的測試 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; }
上傳時間： 2018-05-09
上傳用戶：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.
標簽： decomposition virtual control
上傳時間： 2019-09-04
上傳用戶：txb96
