有如下程序段#include "stdio.h"typedef struct aa{ int a; struct aa *next;} M;void set（M *k,int i,int *b）{ int j,d=0; for（j=1;j<i;j++） { k［j-1］.next=&k［j］; k［j-1］.a=b［d++］; } k［j］.a=b［d］;}main(){ M k［5］,*p; int d［5］={23,34,45,56,67}; set（k,5,d）; p=k+1; printf（"%d＼n",________）;}要输出45，则在下画线处应填入的选项是

A．p->next->a

B．++p->a

C．（*p）．a++

D．p++ ->a

[单选题]有如下程序段#include "stdio.h"typedef union{ long x［2］; int y［4］; char z［8］;}atx;typedef struct aa { long x［2］; int y［4］; char z［8］;} stx;main(){ printf("union=%d,struct aa=%d＼n",sizeof(atx),sizeof(stx));}则程序执行后输出的结果是A．union=8,struct aa=8 B．union=8,str

[单选题]有如下程序段#include "stdio.h"#include "string.h"#define N 10#define M 10char *find(char(*a)［M］,int n){ char *q;int i; q=a［0］; for(i=0;i<n;i++) if(strcmp(a［i］,q)<0)q=a［i］; return q;}main(){ char s［N］［M］={"tomeetme","you","and","he","chin

[单选题]现有如下程序段#include "stdio.h"#include "string.h"main(){ char a[]="acfijk"; /*这里是有序的字符序列*/char b[]="befijklqswz"; /*这里是有序的字符序列*/char c[80],*p;int i=0,j=0,k=0;while(a[i]!=′/0′&&b[j]!= ′/0′){ if(a[i]<b[j])c[k++]=a[i++];else if(a[i]>b[j

[单选题]现有如下程序段#include "stdio.h"#include "string.h"main(){ char a[]="acfijk"; /*这里是有序的字符序列*/char b[]="befijklqswz"; /*这里是有序的字符序列*/char c[80],*p;int i=0,j=0,k=0;while(a[i]!=′/0′&&b[j]!= ′/0′){ if(a[i]<b[j])c[k++]=a[i++];else if(a[i]>b[j

[单选题]现有如下程序段#include "stdio.h"main(){ int a[5][6]={23,3,65,21,6,78,28,5,67,25,435,76,8,22,45,7,8,34,6,78,32,4,5,67,4,21,1};int i=0,j=5;printf("%d/n",*(&a[0][0]+2*i+j-2));}则程序的输出结果为A．21B．78C．23D．28

[单选题]现有如下程序段#include "stdio.h"main(){ int a[5][6]={23,3,65,21,6,78,28,5,67,25,435,76,8,22,45,7,8,34,6,78,32,4,5,67,4,21,1};int i=0,j=5;printf("%d/n",*(&a[0][0]+2*i+j-2));}则程序的输出结果为A．21B．78C．23D．28

[单选题]现有如下程序段#include "stdio.h"main( ){ int a[5][6]={23,3,65,21,6,78,28,5,67,25,435,76,8,22,45,7,8,34,6,78,32,4,5,67,4,21,1};int i=0,j=5;printf("%d/n",*(&a[0][0]+2*i+j-2));}则程序的输出结果为A．21B．78C．23D．28

[单选题]现有如下程序段#include "stdio.h"main(){ int k[30]={12,324,45,6,768,98,21,34,453,456};int count=0,i=0;while(k[i]){ if(k[i]%2==0||k[i]%5==0)count++;i++; }printf("%d,%d/n",count,i);}则程序段的输出结果为A．7,8B．8,8C．7,10D．8,10

[单选题]现有如下程序段#include "stdio.h"main(){ int k[30]={12,324,45,6,768,98,21,34,453,456};int count=0,i=0;while(k[i]){ if(k[i]%2==0||k[i]%5==0)count++;i++; }printf("%d,%d/n",count,i);}则程序段的输出结果为A．7,8B．8,8C．7,10D．8,10

[单选题]有如下程序段#include "stdio.h"void fun(int *a,int *b,int *c,int *d,int *e){ int i,j,k,m; for(i=0;i< *a;i++) for(j=0;j< *b;j++)  for(k=0;k<*c;k++)  for(m=0;m< *d;m++)  ++*e;}main(){ int a=10,b=10,c=10,d=10,e=0; fun(&a,&b,&c,&