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])c[k++]=b[j++];

else { c[k++]=b[j++];

i++;}}

while(a[i]==′/0′&&b[j]!= ′/0′)

c[k++]=b[j++];

while(a[i]!=′/0′&&b[j]== ′/0′)

c[k++]=a[i++];

c[k]=′/0′;

puts(c);}

则输出结果是

A．acfijkbefijklqswz

B．abceffiijjkklqswz

C．befijklqswz acfijk

D．abcefijklqswz

[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]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"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]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" #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]!二'/0'&&b[j]!='/0') { if(a[i]＜b[j])c[k++]=a[i++]； else iA．acfij

[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

[单选题]现有如下程序段#include "stdio.h"int fun（int k,int *m）{if（k%3）*m=k*k;else *m=k/3;}main（）{ int （*p）（int,int *）,m;p=fun;（*p）（78, &m）;printf（ "%d＼n",m）;}则程序段的输出结果为A．24B．25C．26D．27