服务器server0处于不兼容的状态怎么办

你上传后的图片展示应该用上传后的网络路径，比如："http://www.yourdamain.com/photo/20100120091326.jpg"，而不是用"E:webjob0231personpersonPhoto20100120091326.jpg"这样的本地路径。因为在网络应用程序对本地图片没有访问权限。 -----------------------------------------------------------当你保存的时候，相当于服务端的内部逻辑，所以用FileUpload1.PostedFile.SaveAs(Server.MapPath("../person/personPhoto/" + filename));是没错的。但当从客户端打开网络应用程序时，调用的路径是相对于客户端的，比如"E:webjob0231personpersonPhoto20100120091326.jpg"，这被认为是客户端的路径，和你直接在浏览器上键入这个地址是一样的效果，打开的是本地文件而不是服务端文件。所以在面向客户端的时候，比如展示图片，必须用网络路径。

DA1是你定义的变量，变量是不可以用HIGH和LOW操作的！

歌曲名:Is That You?歌手:Bill Frisell专辑:Is That You?Bow Wow - Is That You (P.Y.T)Im Looking for thatpretty young thangup in the spot, looking hot playin` no gamesAnd she steady tryina get it crunk right from the jumpand she playin` no games.that you Shawty? (yea thats me)Did you come here to party? (Yes indeed)Ha But could you drop it down lowwhen its time to get lowif so baby girl lets go.(Come and get it boy)They say I act old, but I got that old man gameCause I stay tryina find me a pretty young thangthat I call anytime if she ready to hangand if sun"it jump off she be ready to bangI like `em fresh outta high schooljust about to graduate short shirtsed`day tryina show they little shapelow cut jump jeans lugz on hea feetneighborhood girl but she love the streetsIn a party she da main one doin dat right thurrEv"time hea jam drop hands up in the airr.And she aint thinkin you, she aint thinkin bout hea hurrAnd if you ballin homie she dont even really currShe got her mind set on what she wanna doAnd she aint come alone she got her whole little crew.Her P.Y.T"s that"ll make ya head spinEry nigga in the club tryina get inIm Looking for thatpretty young thangup in the spot, looking hot playin` no gamesAnd she steady tryina get it crunk right from the jumpand she playin` no games.that you Shawty? (yea thats me)Did you come here to party? (Yes indeed)Ha But could you drop it down lowwhen its time to get lowif so baby girl lets go.(Come and get it boy)They got a whole lot of attituderap is all that they listen tostays in the sweet tryina find somthin newcustom ipods with the rhinestonesnextel camara phones with all the hot ringtones.Light skin til black little PYT"sboricua mami"s that like to eat Mickey D"salways into sumthin like NWAgota little bittie frame but home she dont playYou know a gyrl is the reason that a nigga wanna stuntAnd im weak for em homie I aint even gon frontSee most young nigga"s like em oldernot me spot they spot me like in the summerI run up hit them with that Moss A gamelike "Baby wuz ya phone numba? Were ya live at?Shawdy what part of town? Cuz I just wanna be down!"Im Looking for thatpretty young thangup in the spot, looking hot playin` no gamesAnd she steady tryina get it crunk right from the jumpand she playin` no games.that you Shawty? (yea thats me)Did you come here to party? (Yes indeed)Ha But could you drop it down lowwhen its time to get lowif so baby girl lets go.(Come and get it boy)Pretty Young Thang repeat after mesay Nah nanana "Nah nanana"Nah nanana "Nah nanana"say pretty young thangs repeat after mesay Nah nanana "Nah nanana"Nah nanana "Nah nanana"Im Looking for thatpretty young thangup in the spot, looking hot playin` no gamesAnd she steady tryina get it crunk right from the jumpand she playin` no games.that you Shawty? (yea thats me)Did you come here to party? (Yes indeed)Ha But could you drop it down lowwhen its time to get lowif so baby girl lets go.(Come and get it boy)http://music.baidu.com/song/1061230

对于这个问题，不需要存到数组。按照如下流程即可：1 输入一个字符，对该字符进行判断：a) 如果是数字，则数字累加器加一。b) 如果是字母，则字母累加器加一。c)如果是换行，则结束统计(以换行为结束符。如需其他结束符，根据需要更改判断)。2输出结果。代码：#include <stdio.h>int main(){int c, n, i;c=n=0;while(1){i = getchar();if(i>="0" && i <= "9") n++;else if((i>="a" && i <= "z")||(i>="A" && i <= "Z"))c++;else if(c==" ") break;}printf("数字%d个，字母%d个 ", n,c);return 0;}

struct s {int a,b;} data[2]={10,100,20,200}; 定义了一个数组，data[1]中的a未20b为200然后p=a[1]，把a[1]的值赋给p

#include<stdio.h>int main(){ int date=0,da=0,xiao=0; char c; c=getchar(); while(c!="!") { if(c>="a"&&c<="z") xiao++; if(c>="A"&&c<="Z") da++; if(c>="0" &&c<="9") date++; c=getchar(); // 输入多个字符} printf("xiao=%d,da=%d,date=%d ",xiao,da,date); return 0;}

/*要求:1.集合元素不超过100个。2.集合用数组表示(数组预先定义大一些)。功能有:1.程序初始运行时任意输入一批初始数放入集合(可少可多，不少于5个，不超过30个)。2对该批数排序存放。程序完成前两步后，通过循环可以不断选择下面的某项功能执行(用提示语句列出菜单，通过读入某变量值，switch判断该变量，如接受到1表示执行第一项，依次类推，某项功能执行完可以返回菜单，也可退出系统):3.任意插入一个数到排好序的集合中，保持集合依然有序。4.在集合中删除一个元素。5.查询某个值是否在集合中(该数由键盘输入)。6.求出该数据集合的最大值和最小值及所在位置。7.找出数据集合中的所有素数。*/#include<iostream.h>#include<windows.h>/*charmenu[][100]={" **************************************","请选择操作:","1:输出现有数据;","2:插入一个数据;","3:删除一个数据;","4:查找指定数据;","5:找出数组中最大值","6:找出数组中最小值;","0:退出","*************************************** "};*/charmenu[]=" *************************************** 请选择操作: 1:输出现有数据; 2:插入一个数据; 3:删除一个数据; 4:查找指定数据; 5:找出数组中最大值; 6:找出数组中最小值; 0:退出。 *************************************** ";classArry{private:intarry[110];intnum;public:Arry();//用随机产生的数初始化数组并排序voidInsert(intia);//插入一个数voidShow();//输出现有数组voidDelete(intda);//删除指定的数voidFind(intfa);//查找指定的数voidFindMax();//找出最大数voidFindMin();//找出最小数};Arry::Arry(){num=30;inttemp=0;for(inti=0;i<110;i++)if(i<30)arry[i]=rand()/100;elsearry[i]=0;for(i=0;i<num-1;i++)for(intj=0;j<num-i-1;j++)if(arry[j]>arry[j+1]){temp=arry[j];arry[j]=arry[j+1];arry[j+1]=temp;}}voidArry::Insert(intia){inti=0;if(arry[0]>=ia){for(intj=num;j>0;j--)arry[j]=arry[j-1];arry[0]=ia;num++;}else{while(arry[i]<ia&&i<num)i++;for(intj=num;j>i;j--)arry[j]=arry[j-1];arry[j]=ia;num++;}}voidArry::Show(){cout<<"现有数据为: ";for(inti=0;i<num;i++){cout<<arry[i]<<" ";if((i+1)%5==0)cout<<endl;}cout<<endl;}voidArry::Find(intfa){intf=0;for(inti=0;i<num;i++)if(arry[i]==fa){f=1;cout<<"数组中第"<<i+1<<"个数是";cout<<arry[i]<<endl;}if(f==0)cout<<"数组中没有找到指定的数!"<<endl;}voidArry::Delete(intda){ints=0;for(inti=0;i<num;i++)if(arry[i]==da){for(intj=i;j<num-1;j++)arry[j]=arry[j+1];arry[j]=0;i--;num--;s++;}if(s)cout<<"删除了"<<s<<"个数据!"<<endl;elsecout<<"数组中没有找到指定的数!"<<endl;}voidArry::FindMax(){intp=0;intmax=arry[0];for(inti=1;i<num;i++)if(arry[i]>=max){max=arry[i];p=i;}cout<<"数组中最大数据是:"<<max<<endl;cout<<"在数组中的位置是第"<<p+1<<"个"<<endl;}voidArry::FindMin(){intp=0;intmin=arry[0];for(inti=1;i<num;i++)if(arry[i]<=min){min=arry[i];p=i;}cout<<"数组中最小数据是:"<<min<<endl;cout<<"在数组中的位置是第"<<p+1<<"个"<<endl;}voidmain(){Arrya;intchoice=0;intdata=0;while(1){//for(inti=0;i<10;i++)//cout<<menu[i]<<endl;cout<<menu;cin>>choice;switch(choice){case0:exit(1);break;case1:a.Show();break;case2:cout<<"输入要插入的数据:";cin>>data;a.Insert(data);break;case3:cout<<"输入要删除的数据:";cin>>data;a.Delete(data);break;case4:cout<<"输入要查找的数据:";cin>>data;a.Find(data);break;case5:a.FindMax();break;case6:a.FindMin();break;default:cout<<"操作错误! ";break;}}}

以下是一个简单的 C 语言程序，可以用于输入十个数，并求出它们的最大值、最小值和平均值：```c#include <stdio.h>int main() { int i; double sum = 0.0, max, min, num; // 输入第一个数，并将其作为最大值和最小值的初始值 printf("请输入第1个数: "); scanf("%lf", &num); max = min = num; sum += num; // 循环输入剩余9个数，并更新最大值、最小值和总和 for (i = 2; i <= 10; i++) { printf("请输入第%d个数: ", i); scanf("%lf", &num); if (num > max) { max = num; } if (num < min) { min = num; } sum += num; } // 输出结果 printf("最大值为: %.2f ", max); printf("最小值为: %.2f ", min); printf("平均值为: %.2f ", sum / 10); return 0;}```在这个程序中，我们使用了一个 `for` 循环来输入十个数，并分别更新最大值、最小值和总和。需要注意的是，我们在循环开始时已经输入了第一个数，并将其作为最大值和最小值的初始值。最后，我们输出计算得到的最大值、最小值和平均值。需要注意的是，在计算平均值时，我们使用了总和除以10的方式来求解，因为本程序中输入了恰好十个数。如果输入的数目不是十个，应该相应地调整计算平均值的方式。

没什么问题啊，除了 第7行 // int ;

//你的错误太多，给你个新的#include<stdio.h>#include<stdlib.h>void da(int a[][4],int n){int i,j,row=0,col=0,max=a[0][0];for(i=0;i<n;i++)for(j=0;j<4;j++){if(a[i][j]>max){max=a[i][j];row=i;col=j;}}printf("最大值：a[%d][%d]=%d ",row,col,max); }int main(){int i,j,n,a[10][4];scanf("%d",&n);for(i=0;i<n;i++)for(j=0;j<4;j++)scanf("%d",&a[i][j]);da(a,n);return 0; }

这是一个仿真实例，可以参考一下，试试。

正确答案，可见：http://zhidao.baidu.com/question/151649564

AD呼吸灯是什么？如果需要51输出PWM的话可以使用定时器，里89C52为例推荐是定时器T2作为发生器，可以作为ADC0809这样AD的工作频率

挺简单啊，时间一到就进入中断，中断完再回到主程序

#include<stdio.h>#include<stdlib.h>#define NULL 0void main(){FILE *fp1=NULL,*fp2=NULL;int xiao=0,da=0,shu=0,all[26],i;char ch;fp1=fopen("你要读取的文本名字.txt","r");//要把你要读取的文本放在程序所在的地方fp2=fopen("你想要保存位的文本的名字.txt","w");if(fp1==NULL) { printf("Can not open "); exit(0); }if(fp2==NULL) { printf("Can not open "); exit(0); }for(i=0;i<26;i++) all[i]=0;ch=fgetc(fp1);while(ch!=EOF) { if((ch>=65)&&(ch<=90)) { da++; all[ch-65]++; } else if((ch>=97)&&(ch<=122)) { xiao++; all[ch-97]++; } else if((ch>=48)&&(ch<=57)) { shu++; } ch=fgetc(fp1); }printf("小写字母个数为%d 大写字母个数为%d 数字个数为%d ",xiao,da,shu);fprintf(fp2,"小写字母个数为%d 大写字母个数为%d 数字个数为%d ",xiao,da,shu);for(i=0;i<26;i++)fprintf(fp2,"%c=%d ",i+65,all[i]);fclose(fp1);fclose(fp2);}vc++6.0运行成功

霹雳英雄布袋戏、不谢

#include<stdio.h>void main(){ int i,max,min,da,xiao,a[10]; printf("请输入任意10个整数 "); for(i=0;i<10;i++) { scanf("%d",&a[i]); } max=a[0]; min=a[0]; for(i=1;i<10;i++) { if(a[i]>max) { max=a[i]; da=i;} if(a[i]<min){ min=a[i]; xiao=i;} } a[da]=a[0]; a[xiao]=a[9]; a[0]=max; a[9]=min; for(i=0;i<10;i++) printf("%d ",a[i]);}

#include<iostream>#define max 100using namespace std;class bank{ int top; char date[max][10]; float money[max]; float rest[max]; static float sum;public: bank(){top=0;} void bankin(char d[],float m) { // 数组的名称即可作为指针使用 strcpy(date[top],d); money[top]=m; sum=m+sum; rest[top]=sum; top++; } void bankout(char d[],float m) { strcpy(date[top],d); money[top]=-m; sum=sum-m; rest[top]=sum; top++; } void disp() { int i; printf(" 日期 存入 取出 余额 "); for(i=0;i<top;i++) { printf("%8s ",date[i]); if(money[i]<0) printf("%6f ",-money[i]); else printf("%f ",money[i]); printf("%6d ",rest[i]); } }};float bank::sum=0;void main(){ bank obj; obj.bankin("2002.2.5",1000); obj.bankin("2001.3.2",2000); obj.bankout("2001.4.1",500); obj.bankout("2001.4.5",800); obj.disp();}

h> void fun(int a[][34],int n) { int i,j; for(i=0;i<n;i++) for(j=0;j<=i;j++) { if(i==j) a[i][j]=1; else if(j==0) a[i][j]=1; else a[i][j]=a[i-1][j-1]+a[i-1][j]; } } int main() { int i,j,k,n; int a[34][34]; k=1; while(scanf("%d",&n)==1) // 输入你要显示的行数，如1,2,3,4,... { fun(a,n); printf("Case %d: ",k++); for(i=0;i<n;i++) { for (j = 0; j < n - i; j ++) printf(" "); for(j=0;j<=i;j++) { if(j==i) printf("%d",a[i][j]); else printf("%d ",a[i][j]); } printf(" "); } printf(" "); } return 0; }第2个:#include<stdio.h>struct d{ int year; int month; int day;}data;int main(){ int m(struct d da); int n; printf("输入日期:"); scanf("%d%d%d",&data.year,&data.month,&data.day); n=m(data); printf("该日期是%d年第%d天",data.year,n); return 0;}int m(struct d da){ if(da.month==1) return da.day; int i,sum=0,mo[12]=; for(i=0;i<da.month-1;i++) { sum=sum+mo[i]; } sum+=da.day; if(da.year%400==0||da.year%4==0&&da.year%100!=0) if(da.month>2) sum++; return sum;}

#include<iostream> using namespace std; template <typename T>class List;template <typename T>class Node { friend class List<T>;private: T data; Node<T>* next;public: Node(){data=NULL;next=NULL;} Node(T& da) {data =da;next=NULL;} Node(T& da,Node<T>* ne) {data=da;next = ne};};template <typename T>class List{private: int length; Node<T>* head;//头结点,其next节点为第一个节点 Node<T>* last;//指向最后一个结点的指针public: List(); void insert (T& data); void remove (T data); int getLength() const{return length;}; void print () const; void reverse();};template <typename T>List<T>::List(){ length = 0; head = new Node<T>(); last = head;}template <typename T>void List<T>::insert(T &data){ last = last ->next =new Node<T>(data); length++;}template <typename T>void List<T>::remove(T data){ Node<T>* current = head; while (current->next!=NULL) { if(current->next->data==data) { Node<T>* temp = current->next; current->next = current->next->next; delete temp; length--; } else current = current->next; }}template <typename T>void List<T>::print() const{ Node<T>* current =head; while (current->next!=NULL) { current = current->next; cout<<current->data<<" "; }}template <typename T>void List<T>::reverse(){ Node<T>* temp1;//利用三个临时指针降List中每个结点的next倒向 Node<T>* temp2; Node<T>* temp3; last = head->next; temp1 = head; temp2 = temp1->next; while(temp2!=NULL) { temp3 = temp2->next; temp2->next = temp1; temp1 = temp2; temp2 = temp3; }//倒向完毕 last->next = NULL;//重新设置head和last head->next = temp1;}int main ()//测试代码{ List<int> one; int i1=34,i2=89,i3=232; one.insert(i1); one.insert(i1); one.insert(i1); one.insert(i1); one.insert(i2); one.insert(i3); one.print(); cout<<endl; one.reverse(); one.print(); cout<<endl; one.remove(34); one.print();}我用的是vc2008,应该没问题,都按照要求了

#include <stdio.h>#include <stdlib.h>#include <string.h>struct A{ int a; char b[10]; double c;};struct A f(struct A t);void main(){ struct A a = {1001,"ZhangDa",1098.0}; a = f(a); printf("%d,%s,%6.1f ",a.a,a.b,a.c);}struct A f(struct A t){ t.a=1002; strcpy(t.b,"ChangRong"); t.c=1202.0; return t;}

#include<iostream.h>static int dys[]={31,28,31,30,31,30,31,31,30,31,30,31};class date{ int yr,mo,da;public: date(int y,int m,int d) { yr=y; mo=m; da=d; } date() {} void disp() { cout<<yr<<"年"<<mo<<"月"<<da<<"日"<<endl; } date operator + (int day) { date dt=*this; day+=dt.da; while (day>dys[dt.mo-1]) { day-=dys[dt.mo-1]; dt.mo++; if (dt.mo==13) { dt.mo==1; dt.yr++; } } dt.da=day; return dt; }};void main(){ int y,m,d,a,b; cout<<"输入年月日:"; cin>>y>>m>>d; date d1(y,m,d),d2,d3; d1.disp(); cout<<"输入要加的天数（整数）:"; cin>>a; d2=d1+a; d2.disp();}

Aseem不用于进行时态和被动语态中

双击led，在mode type的属性里调成digital

#include <stdio.h>void main() { int x=8; for(;x>0;x--) { if(x%3) { printf("%d ",x--); } }}不知道你说什么意思，帮你完整了下

你的空间没释放！！free(s);free(root);

首先，判断（a++&&b++）这个条件，这个条件中又首先判断a++，得到结果为假（由于是后置自增运算符，所以a在判断的时候为0，判断之后为1）。&&运算中，只要有一个条件为假，则整个条件为假，所以程序不再判断b++这个条件，并且将(a++&&b++)这个条件判定为假。所以程序执行冒号后面的d++。所以a=1,b=0,c=0,d=1。所以答案选D

应付考试，首选谭浩强的书啦~

若有高手，望详细解答，小弟也很小知道，谢谢

不加while(1)则程序会不停地循环执行整个main()函数，加了的话就会停在while(1)处了。

你是沈师软件大一的吧???TC 2.0 /* welcome to use the WanNianLi system! Copyright @ 2005 KongXinCai All rights reserved!:):):)*/ #include<stdio.h> #include<stdlib.h> char* month_str[]={"January","February","March","April","May","June","July","August","September","October","November","December"}; char* week[]={"Sunday","Monday","Tuesday","Wednesday","Thursday","Friday","Saturday"}; int IsLeapYear(int year) /*find out the year is leap year or not*/ { if((year%4==0&&year%100!=0)||(year%400==0)) return 1; else return 0; } int month_day(int year,int month) { int mon_day[]={31,28,31,30,31,30,31,31,30,31,30,31}; if(IsLeapYear(year)&&month==2) return 29; else return(mon_day[month-1]); } int DaySearch(int year,int month,int day) /*search what day this day is*/ { int c=0; float s; int m; for(m=1;m<month;m++) c=c+month_day(year,m); c=c+day; s=year-1+(float)(year-1)/4+(float)(year-1)/100+(float)(year-1)/400-40+c; return ((int)s%7); } int PrintAllYear(int year)/*print the all year*/ { int temp; int i,j; printf(" %d Calander ",year); for(i=1;i<=12;i++) { printf(" %s(%d) ",month_str[i-1],i); printf("0 1 2 3 4 5 6 "); printf("S M T W T F S "); temp=DaySearch(year,i,1); for(j=1;j<=month_day(year,i)+temp;j++) { if(j-temp<=0) printf(" "); else if(j-temp<10) printf("%d ",j-temp); else printf("%d ",j-temp); if(j%7==0) printf(" "); } } return 0; } int main() { int option,da; char ch; int year,month,day; printf("Copyright @ 2005 TianQian All rights reserved!:):):)"); printf(" Welcome to use the WanNianLi system! "); while(1) { printf(" Please select the service you need: "); printf(" 1 Search what day the day is"); printf(" 2 Search whether the year is leap year or not"); printf(" 3 Print the calander of the whole year"); printf(" 4 Exit "); scanf("%d",&option); switch(option) { case 1: while(1) { printf(" Please input the year,month and day(XXXX,XX,XX):"); scanf("%d,%d,%d,%c",&year,&month,&day); da=DaySearch(year,month,day); printf(" %d-%d-%d is %s,do you want to continue?(Y/N)",year,month,day,week[da]); fflush(stdin); scanf("%c",&ch); if(ch=="N"||ch=="n") break; } break; case 2: while(1) { printf(" Please input the year which needs searched?(XXXX)"); scanf("%d",&year); if(IsLeapYear(year)) printf(" %d is Leap year,do you want to continue?(Y/N)",year); else printf(" %d is not Leap year,do you want to continue(Y/N)?",year); fflush(stdin); scanf("%c",&ch); if(ch=="N"||ch=="n") break; } break; case 3: while(1) { printf(" Please input the year which needs printed(XXXX)"); scanf("%d",&year); PrintAllYear(year); printf(" Do you want to continue to print(Y/N)?"); fflush(stdin); scanf("%c",&ch); if(ch=="N"||ch=="n") break; } break; case 4: fflush(stdin); printf("Are you sure?(Y/N)"); scanf("%c",&ch); if(ch=="Y"||ch=="y") exit(0); break; default: printf(" Error:Sorry,there is no this service now! "); break; } } return 0; }

先排序，再求差

typedef double (*point) [2]; point read(int n); 这样如何？

以下叙述不正确的是（）x0dA:一个C程序可由一个或多个main函数组成x0dB:一个C源程序必须包含一个main函数x0dC:C程序的基本组成单位是函数x0dD:在C程序中,注释说明只能位于一条语句的后面5

要DOS下的图形界面的???C++/C#可以不?C太麻烦了 懒得写

public static void main(string[] args){}

#include <stdio.h>#include <stdlib.h>int main(){ int n,sum,i; sum=0; for(i=0; i<20; i++) { n=1+i*2; if(i%2== 1) n=-n; sum+=n; } printf("result：%d",sum); return 0;}

你好！你选的那个DA芯片？，是做实物还是仿真就行？

input函数中，需要改为char a[10][80];int i;

输入用gets,然后分别判断各字符在"0"-"9"之间,还是在"a"-"z"或"A"-"Z"之间

现在不动手，将来想做这一行，还是要动手，自己动动手吧，也不难。

写一个程序即可。程序详情：#include "stdio.h"main(){int a,b,c;scanf("%d%d%d",&a,&b,&c);if(a>b&&a>c){if(b>c) printf("%d,%d,%d",a,b,c);else printf("%d,%d,%d",a,c,b);}if(b>c&&b>a){if(a>c) printf("%d,%d,%d",b,a,c);else printf("%d,%d,%d",b,c,a);}if(c>a&&c>b){if(a>b) printf("%d,%d,%d",c,a,b);else printf("%d,%d,%d",c,b,a);}getchar();}

已发

#include<stdio.h>void fun2(char a1,char a2,char *a3,int b1,int b2,int *b3,float c1,float c2,float *c3){*a3=a1+a2;*b3=b1+b2;*c3=c1+c2;}void fun1(void){char a1=1,a2=2,a3;int b1=100,b2=300,b3;float c1=5.1,c2=6.8,c3;fun2(a1,a2,&a3,b1,b2,&b3,c1,c2,&c3);printf("a3=%d b3=%d c3=%f ",a3,b3,c3);}int main(){fun1();return 0;}

#include<stdio.h>int main(){char c;int letters=0,space=0,digit=0,other=0;printf("请输入一段字符 ");while((c=getchar())!=" "){if(c>="a"&&c<="z"||c>="A"&&c<="Z")letters++;else if(c==" ")space++;else if(c>="0"&&c<="9")digit++;elseother++;}printf("字母数:%d 空格数:%d 数字数:%d 其他字符数:%d ",letters,space,digit,other);return 0;}如果没有问题请采纳 谢谢

简单的看了下，你代码写的也太不规范了吧？中西结合啊，你以为中西结合代码就很牛B吗？还是全英文吧。Fill方法有11个重载版本的，但是也不是很复杂，因为我们用的时候也就那么一两个。Fill(DataSet）和Fill(dataset ,tablename)。例如Da.Fill(objDataSet)，或者Da.Fill(objDataSet,"Users")。我经常用第二种。你代码中的PasswordTextBox.Text是怎么回事？恕我眼拙！

你这是考试题，还是自己看书吧，书上写的很清楚…

显然是对的嘛!!!

#include <stdio.h>enum something{ aa,bb,cc,dd,ee,ff,gg} ;void main(){enum something x[2][3]; int i,j;x[0][0]=dd;x[0][1]=ee;x[0][2]=ff;x[1][0]=aa;x[1][1]=bb;x[1][2]=cc;for (j=0;j<2;j++){for (i=0;i<3;i++) printf("%d ",x[j][i]);printf(" ");}}-------------------------------C++ 头文件换#include<iostream>using namespace std;输出用for (j=0;j<2;j++){for (i=0;i<3;i++) cout << x[j][i] << " ";cout << endl;}

/*开发环境：turboc2.0模拟时钟转动程序代码*/#include"graphics.h"#include"math.h"#include"dos.h"#definepi3.1415926#defineX(a,b,c)x=a*cos(b*c*pi/180-pi/2)+300#defineY(a,b,c)y=a*sin(b*c*pi/180-pi/2)+240#defined(a,b,c)X(a,b,c);Y(a,b,c);line(300,240,x,y)voidinit()/*划时钟边框函数*/{inti,l,x1,x2,y1,y2;setbkcolor(1);circle(300,240,200);circle(300,240,205);circle(300,240,5);for(i=0;i<60;i++)/*划钟点上的短线*/{if(i%5==0)l=15;elsel=5;x1=200*sin(i*6*pi/180)+300;y1=200*cos(i*6*pi/180)+240;x2=(200-l)*sin(i*6*pi/180)+300;y2=(200-l)*cos(i*6*pi/180)+240;line(x1,y1,x2,y2);}}main(){intx,y,i,k=1;intgdriver=9,gmode=2;unsignedcharh,m,s;into,p,q;floatn;structtimet[1];structdated[1];initgraph(&gdriver,&gmode,"c:\tc");initgraph(&gdriver,&gmode,"c:\tc");for(i=0;i<=6;i++){settextstyle(TRIPLEX_FONT,HORIZ_DIR,i);/*控制输出字符的字体，方向，大小*/cleardevice();settextjustify(1,1);/*在指定坐标上输出字符串*/outtextxy(300,80,"12");outtextxy(300,390,"6");outtextxy(140,230,"9");outtextxy(460,230,"3");outtextxy(380,100,"1");outtextxy(220,100,"11");outtextxy(430,160,"2");outtextxy(430,310,"4");outtextxy(380,370,"5");outtextxy(220,370,"7");outtextxy(160,160,"10");outtextxy(160,310,"8");}init();setwritemode(1);/*设置画线的输出模式*/if(k!=0){getdate(d);/*获得系统日期函数*/o=d[0].da_year;p=d[0].da_mon;q=d[0].da_day;gettime(t);/*获得系统时间函数*/h=t[0].ti_hour;m=t[0].ti_min;s=t[0].ti_sec;}setcolor(7);/*设置时针颜色*/n=(float)h+(float)m/60;d(150,n,30);/*画出时针*/setcolor(14);/*设置分针颜色*/d(170,m,6);/*画出分针*/setcolor(4);/*设置秒针颜色*/d(190,s,6);/*画出秒针*/while(!kbhit())/*控制程序按下任意键退出*/{while(t[0].ti_sec==s)gettime(t);gotoxy(44,18);/*使光标移动到指定坐标*/printf("");/*退格，使表示时间的字符串不断变化*/sound(400);/*按给定的频率打开PC扬声器*/delay(70);/*中断程序的执行，时间为70毫秒*/sound(200);delay(30);nosound();/*按给定的频率关闭PC扬声器*/setcolor(4);d(190,s,6);s=t[0].ti_sec;d(190,s,6);if(t[0].ti_min!=m){setcolor(14);d(170,m,6);m=t[0].ti_min;d(170,m,6);}if(t[0].ti_hour!=h){setcolor(7);d(150,h,30);h=t[0].ti_hour;d(150,h,30);sound(1000);delay(240);nosound();delay(140);sound(2000);delay(240);nosound();}if(s<10)/*用字符的形式输出时间*/{if(m<10)printf("%u:0%u:0%u",h,m,s);elseprintf("%u:%u:0%u",h,m,s);}else{if(m<10)printf("%u:0%u:%u",h,m,s);elseprintf("%u:%u:%u",h,m,s);}gotoxy(34,19);/*在指定坐标上输出日期*/printf("%d年%d月%d日",o,p,q);printf("");}getch();closegraph();}