1、閱讀程序模擬題1.下面程序執行結果是#includevoid swap(int *,int *);void main()int a=5, b=7, *ptr1, *ptr2;ptr仁&a, ptr2=&b; swap (ptr1,ptr2); printf(“ *ptr1=%d,*ptr2=%di” ,*ptr1,*ptr2);printf( “ a=%d,b=%nd ,a,b);void swap (int *p1, *p2)int p;p=*p1;*p1=*p2; *p2=p;*ptr1=7,*ptr2=5a=7,b=52.輸入 we are students ,下面程序運行結果 3#in

2、clude void main() char string81; int i,num=0,word=0; char c; gets(string); for(i=0;(c=stringi)!=O;i+) if(c= ) word=0;else if(word=0) word=1; num+; printf( %dn,num);3.下面程序的運行結果是 245,760,11973#include void inv(int x, int n); void main() int i,a1O=3,7,9,11,0,6,7,5,4,2; inv(a,10);printf(The array has be

3、en 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;iv=m;i+) j=n-1-i;t=xi; xi=xj; xj=t;4 下面程序的運行結果是 #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=%snst

4、ring_b=%sn,a,b);void copy_string(char *from,char *to) for(;*from!=0;from+,to+)*to=*from;*to=0; string_a= I am a teacher. string_b= I am a teacher.5 下面程序的運行的功能是 #include#includevoid sort(char *name,int );void print(char *name,int);void main()char *name=Follow me,BASIC,Great Wall,FORTRAN,Computer ; in

5、t n=5;sort(name,n);print(name,n); void sort(char *name,int n) char *temp;int i,j,k;for(i=0;i0) k=j;if(k!=i) temp=namei; namei=namek; namek=temp;void print(char *name,int n)int i; for(i=0;ivn;i+) puts(namei);ASICComputerFORTRANFollow meGreat Wall6.輸入5回車后，下面程序的執行結果是120#include long int fac(int ); void

6、 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;else仁 fac(m-1)*m;return f;7 .下面程序執行的結果是： 輸入如下數據：LiZhangLiLiWangZhangWangZhang#includestruct personchar name20;int count;leader3= Li,0, Zhang,0, Wang,0;void main() int i,j;char

7、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); Li:3Zhang:3Wang:2&輸入17,下面程序執行的結果是 #include#includevoid main()int m,i,k;scanf(%d,&m); k=sqrt(m); for(i=2;iv=k;i+) if(m%i=O)

8、break;if(ik)printf(%d is a prime numbern,m); elseprintf(%d is not a prime numbern,m);17 is a prime number9 下面程序執行的結果是 #in clude#in cludevoid mai n()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;elseaij+；break;case 1: aij+=2; break;case 2: aij+=3;printf(%d , %d ,

9、%d,a11, a12, a20);將數組定義改為：int a33=0,則結果為：1, 2, 310.寫出下面程序的運行結果：8, 17#in cludeint func (in t ,i nt);void main ()int k=4,m=1,p1,p2;p1=fu nc(k,m);p2=fu nc(k,m);printf( %d,%dn ”,p1,p2);int func (int a,i nt b)static int m=0,i=2;i+=m+1;m=i+a+b;return (m);11 以下程序執行后輸出的結果是123, 0#i nclude void mai n()FILE *f

10、p;int i,k=0,n=0;fp=fope n( d1.dat,w);for(i=1;i4;i+)fprin tf(fp, %d,i);fclose(fp);fp=fope n(d1.dat,r);fsca nf(fp, %d%d,&k, &n);prin tf(%d,%dn,k, n);fclose(fp);12以下程序執行后輸出的結果是 轉置#in cludevoid mai n()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+)prin tf

11、(%4d,bij);prin tf(n ”);-3*2513若輸入的值是-75，以下程序的運行結果是 -#in clude#in clude void fun (i nt );void mai n() int n;scan f(%d,&n);if(n 0) pri ntf(-);n=abs (n);fun(n);void fun (i nt x)int k,r;for(k=2;k1) prin tf(*);r=x%k;if(x!=1) pri ntf(%dn,x);14以下程序的運行結果是4#in clude#in cludestruct NODEint num: struct NODE；vo

12、id mai n()struct NODE int sum=0;*next;*P,*q,*r；p=(struct NODE*) malloc(sizeof(struct q=(struct NODE *) malloc(sizeof(struct r=(struct NODE *) malloc(sizeof(struct 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.輸入87653142，下面程序運行的結果是 753J#in cludeNODE); NODE); NODE);void fun (Io ng in t , l ong int *); void mai n()long s, t;scan f(%ld, &s);fun(s, &t);prin tf(%ldn, t);void fun (Io ng int s, l ong int *t)int d;long s1=1;*t=0;while(s0)d=s%10;if(d%2!=0)*t=d*s1+*t;