C语言阅读程序模拟题.doc

阅读程序模拟题1下面程序执行结果是 #includevoid swap(int *,int *);void main() int a=5, b=7, *ptr1, *ptr2;ptr1=&a, ptr2=&b;swap (ptr1,ptr2);printf(“*ptr1=%d,*ptr2=%dn”,*ptr1,*ptr2);printf(“a=%d,b=%dn”,a,b);void swap (int *p1, *p2)int p;p=*p1; *p1=*p2; *p2=p;2输入we are students ，下面程序运行结果 #include void main() char string81; int i,num=0,word=0; char c; gets(string); for(i=0;(c=stringi)!=0;i+) if(c= ) word=0; else if(word=0) word=1; num+; printf( %dn,num);3下面程序的运行结果是 The array has been reverted: 2，4，5，7，6，0，11，9，7，3 #includevoid inv(int x, int n);void main() int i,a10=3,7,9,11,0,6,7,5,4,2; inv(a,10); printf(The array has been reverted:n); for(i=0;i10;i+) printf(%d,ai); printf(n);void inv(int x, int n) int t,i,j,m=(n-1)/2; for(i=0;i=m;i+) j=n-1-i; t=xi; xi=xj; xj=t; 4下面程序的运行结果是 #includevoid copy_string(char *,char *);void main() char *a=I am a teacher.; char *b=You are a student.; copy_string(a,b); printf(nstring_a=%snstring_b=%sn,a,b);void copy_string(char *from,char *to) for(;*from!=0;from+,to+) *to=*from; *to=0;5下面程序的运行的功能是 BASIC#include#includevoid sort(char *name,int );void print(char *name,int);void main() char *name=Follow me,BASIC, Great Wall,FORTRAN,Computer ; int n=5; sort(name,n); print(name,n);void sort(char *name,int n) char *temp; int i,j,k; for(i=0;in-1;i+) k=i; for(j=i+1;j0) k=j; if(k!=i) temp=namei; namei=namek; namek=temp; void print(char *name,int n) int i; for(i=0;in;i+) puts(namei);6. 输入5回车后，下面程序的执行结果是 #includelong int fac(int );void main()int n;scanf(%d,&n);printf(%dn,fac(n);long int fac(int m)long int f;if(m0)printf(data error!n);else if(m=0|m=1)f=1;elsef=fac(m-1)*m;return f;7下面程序执行的结果是： 输入如下数据：Li Zhang Li Li Wang Zhang Wang Zhang#includestruct person char name20; int count;leader3= Li,0, Zhang,0, Wang,0; void main() int i,j; char leader_name20; for(i=1;i=8;i+) scanf(%s,leader_name); for(j=0;j3;j+) if(strcmp(leader_name,)=0) leaderj.count+; for(i=0;i3;i+) printf(%5s:%dn,,leaderi.count);8输入17，下面程序执行的结果是 #include#includevoid main()int m,i,k;scanf(%d,&m);k=sqrt(m);for(i=2;ik)printf(%d is a prime numbern,m);elseprintf(%d is not a prime numbern,m);9下面程序执行的结果是 _-858993459 _-858993458 -858993457 #include#includevoid main()int a33, i, j;for(i=0;i3;i+)for(j=0;j3;j+) switch(abs(i-j) case 0: if(i=0|i=2) aij+=3;else aij+; break;case 1: aij+=2; break;case 2: aij+=3;printf(%d , %d , %d,a11, a12, a20);10.写出下面程序的运行结果： 8，17 #includeint func (int ,int );void main ( ) int k=4,m=1,p1,p2;p1=func(k,m) ; p2=func(k,m) ;printf(“%d,%dn”,p1,p2) ;int func (int a,int b) static int m=0,i=2;i+=m+1;m=i+a+b;return (m);11以下程序执行后输出的结果是 123,0 #include void main( ) FILE *fp; int i,k=0,n=0;fp=fopen(d1.dat,w);for(i=1;i4;i+) fprintf(fp, %d,i);fclose(fp);fp=fopen(d1.dat,r);fscanf(fp, %d%d,&k, &n); printf(%d,%dn,k, n) ;fclose(fp);12以下程序执行后输出的结果是 1 23 45 6 #includevoid main() int a23=1,3,5,2,4,6; int b32,i,j; for(i=0;i2;i+) for(j=0;j3;j+) bji=aij; for(i=0;i3;i+) for(j=0;j2;j+) printf(%4d,bij); printf(n); 13若输入的值是-75，以下程序的运行结果是 - 3* 25 #include#include void fun(int );void main( ) int n; scanf(%d,&n); if(n0) printf(-);n=abs(n);fun(n);void fun(int x) int k,r; for(k=2;k1) printf(*);r=x%k; if(x!=1) printf(%dn,x);14以下程序的运行结果是 4 #include#includestruct NODEint num;struct NODE *next;void main( )struct NODE *p,*q,*r;int sum=0;p=(struct NODE*) malloc(sizeof(struct NODE);q=(struct NODE *) malloc(sizeof(struct NODE);r=(struct NODE *) malloc(sizeof(struct NODE);p-num=1; q-num=2; r-num=3;p-next=q; q-next=r; r-next=NULL;sum+=q-next-num; sum+=p-num;printf(%dn,sum);15输入87653142，下面程序运行的结果是 7531 #includevoid fun(long int , long int *);void main() long s, t; scanf(%ld, &s); fun(s, &t); printf(%ldn, t);void fun(long int s, long int *t) int d; long s1=1; *t=0; while(s0) d=s%10; if(d%2!=0) *t=d*s1+*t; s1*=10; s/=10; 16下面