二级c语言上机专用题库答案Word文件下载.doc
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b:
(a>
c:
___1___)):
((a>
c)?
___2___:
((b>
___3___));
(1)a
(2)a(3)b
16if((ch>
a'
)___1___(ch<
z'
))
(1)&
&
(2)'
A'
(3)ch
17、if(isdigit(*s))sum+=*s-__1__;
(1)48
(2)s++(3)sum
18、f=___1___;
(1)1
(2)-1(3)t
19、if(n>
=len)strcpy(__1__);
(1)t,s
(2)s[i](3)0
20、__1__fun(structstudent*a)
(1)structstudent*
(2)a->
score[i](3)a
21、max=min=___1___;
(1)a[i]
(2)a[j](3)a[j]
22、voidfun(int___1___,intn)
(1)t[][N]
(2)i=0;
i<
n(3)s
23、voidfun(int___1___,intn)
(1)*a
(2)2(3)i+1
24、f=1.0+___1___;
(1)x
(2)n(3)t
25、t=___1___;
(注意和第1题不一样)
for(i=___2___;
n;
(1)1
(2)1(3)i
26、b__1__=10004;
(1)->
sno
(2)->
name(3)&
t
27、k=__1__;
(注意和第49题不一样)
t=__2__;
(1)1
(2)2*i(3)(-1)
28、s=__1__;
(1)0
(2)n(3)(t*t)
29、ch=tolower(__1__);
(1)*s
(2)1(3)k[n]
30、__1__t;
(1)structstudent
(2)n-1
(3)a[i].name,a[j].name
31、if(___1___==0){
(1)a[i]%2
(2)a[j](3)j
32、{t=*n%__1__;
(1)10
(2)0(3)x
33、for(i=0;
i++)(注意有相类似的)
if(len<
=__2__)
(1)N
(2)k(3)ss[i]
34、__1__;
flag=1;
}
(1)n++
(2)0(3)s++
35、t[j]=__1__;
j++;
(1)s[i]
(2)k(3)0
36、for(i=0;
if(strcmp(ss[i],t)==0)
return__2__;
(注意有类似的)
(1)N
(2)i(3)-1
37、lp=__1__;
(1)s
(2)--(注意是2个减号)
(3)return0
38、x=__1__/4;
(1)3.(注意要打上点)
(2)>
(3)(2*i+1)
39、for(i=0;
___1___;
a[i]=a[n-1-___2___];
(注意有类似题)
(1)n/2
(2)i(3)a[n-i-1]
40、b=__1__;
(1)a
(2)b.name(3)score[i]
41、for(___1___;
j<
i;
j++)
(1)j=2
(2)i(3)j
42、i=__1__;
(1)1
(2)s(3)i*10
43、__1__fun(structstudenta)
(1)structstudent
(2)a.name
(3)a.score[i]
44、p=(n%2==0)?
n/2:
n/2+___1___;
(1)1
(2)i(3)a[p+i]
45、if(a[i]%2==___1___)
(1)1
(2)j++(3)j
46、n=__1__;
(1)0
(2)x(3)t++
47、*(t+n)=__1__;
n++;
}
(1)*s
(2)s++(3)n
48、#defineOK(i,t,n)
((___1___%t==0)&
(i/t<
n))
(1)i
(2)t++(3)count
49、k=__1__;
if(strlen(ps[k])<
strlen(__2__))k=j;
(1)i
(2)ps[j](3)tp
50for(j=a[i]*2;
=n;
j+=___1___)
(1)a[i]
(2)a[i](3)0
二、改错题
1.
(1)for(i=1;
=y;
i++)
(2)t=t%1000;
2.
(1)if(t==0)
(2)*zero=count;
3.
(1)*t=0;
(2)if(d%2!
=0)
4.
(1)for(i=2;
=m;
i++)
(2)y-=1.0/i/i;
5.
(1)t=1;
(2)return(2*s);
6.
(1)#defineFU(m,n)((m)/(n))
(2)return(value);
7.
(1)for(i=strlen(t)-1;
i;
i--)
(2)if(t[j]>
t[j+1])
8.
(1)voidfun(longs,long*t)
(2)while(s>
0)
9.
(1)doubler;
(2)while(fabs(m-n)>
0.001)
10.
(1)b[k]=*p;
(2)b[k]='
'
;
k++;
11.
(1)voidfun(char*s,char*t)
(2)t[2*d]=0;
12.
(1)n=strlen(aa);
(2)ch=aa[i];
13.
(1)fun(intx,inty,intz)
(2)returnj;
14.
(1)voidfun(inta[][M],intm)
(2)a[j][k]=(k+1)*(j+1);
15.
(1)c=tolower(c)(注:
c小写);
(2)c=c+5;
16.
(1)voidfun(int*a)
(2)a[j]=a[j-1];
17.
(1)if((k%13==0)||(k%17==0))(注:
双等号)
(2)}(注:
横线去掉)
18.
(1)for(i=1;
=3;
i++)
(2)if(k>
=0&
k<
=6)
19.
(1)if(i%k==0)
=i)
20.
(1)char*fun(char(*sq)[M])
(2)returnsp;
21.
(1)switch(g)
(2)case1:
return1;
case2:
return1;
22.
(1)fun(inta[],intm)
(2)elseif(m>
a[mid])
23.
(1)sum=0;
(2)if((i+1)%5==0)
24.
(1)d=1;
s=0;
(2)d=d/10;
25.
(1)k=i;
(2)c=k;
26.
(1)doublefun(doublea,doublex0)
(2)if(fabs(x1-x0)>
0.00001)
27.
(1)k++;
(2)if(m==k)
28.
(1)for(i=0;
str[i];
(2)if(substr[k+1]==0)
29.
(1)floatfun(intk)
(2)returns;
30.
(1)q=p+i;
(2)while(q>
p)
31.
(1)fun(intn)
(2)if(n==1)
32.
(1)if(n==0)
(2)result*=n--;
33.
(1)intk=0;
(2)while(*p||*q)
34.
(1)t+=s[k];
(2)*aver=ave;
35.
(1)while(*w)
(2)if(*r==*p)
36.
(1)a2=k/10;
(注意是十)
(2)returni;
37.
(1)doublefun(intn)
(2)c=a;
a+=b;
b=c;
38.
(1)doublefun(intn)
(2)s=s+(double)a/b;
39.
(1)n=*p-'
(2)n=n*8+*p-'
40.
(1)s[j++]=s[i];
(2)s[j]=0;
41.
(1)sum=0;
j=0;
(2)if(sum%4==2)
42.
(1)intfun(intn,intxx[][M])
(2)printf("
%d"
xx[i][j]);
43.
(1)if(p==n)return-1;
(2)a[i]=a[i+1];
44.
(1)floatk;
(2)if(*c>
*a)
45.
(1)t=a;
a=b;
b=t;
(2)return(b);
46.
(1)inti,sl;
(2)t[i]=s[sl-i-1];
47.
(1)intj,c=0;
doublexa=0.0;
(2)if(x[j]>
=xa)
48.
(1)intfun(inta,intb,intc)
(2)elsereturn1;
49.
(1)doublefun(intn)
(2)returnsum;
50.
(1)y=1;
(2)d=a-i;
三、编程题
1、inti=0;
doubles=1,a=1;
while(fabs(a)>
0.000001)
{a=a*(0.5-i)*x/(i+1);
s+=a;
i++;
returns;
2、inti;
doubles=0,a=1;
for(i=1;
{a=a/i;
s+=a;
3、while(*p1)p1++;
while(*p2)
{*p1=*p2;
p1++;
p2++;
*p1=0;
4、inti=0,s=0,n=0;
for(i=0;
m;
i++)s+=score[i];
s/=m;
{if(score[i]<
s)
{below[n]=score[i];
n++;
}
returnn;
5、inti=0,k=0;
char*t=a;
while(*t=='
*'
)
{
t++;
t=a;
if(k>
n)
t=a+k-n;
while(*t)
a[i]=*t;
i++;
a[i]='
\0'
6、inti=0,j=0;
while(a[i]=='
)i++;
while(a[i])
{a[j]=a[i];
j++;
a[j]=0;
7、inti=1;
while(ss[i-1]&
ss[i])
{if(ss[i]>
ss[i]<
)ss[i]+='
-'
i+=2;
8、inti,j;
STRECstrTemp;
for(i=0;
N-1;
for(j=i+1;
j<
N;
j++)
if(a[i].s<
a[j].s)
{
strTemp=a[i];
a[i]=a[j];
a[j]=strTemp;
}
9、voidfun(char*a,char*h,char*p)
{a=h;
while(h!
=p)
{if(*h!
{*a=*h;
a++;
h++;
while(*p)
{*a=*p;
a++;
p++;
*a=0;
10、inti=0;
doubles=0;
while(h->
next)
{h=h->
next;
s+=h->
s;
s/=i;
11、inti=0,j=0;
while(s[i])
{if(i%2==1&
s[i]%2==1)
{t[j]=s[i];
j++;
t[j]=0;
12、doubles=0;
if(s<
h->
s)s=h->
13、inti;
n-h-e;
i++)a[i]=a[i+h];
a[i]=0;
14
*c=a/10*10+a%10*1000+b/10+b%10*100;
15、inti;
doubles=0,a=0;
{a+=sqrt(i);
16、inti,s=0;
for(i=2;
{if(n%i==0)s+=i;
17、inti,s=0;
for(i=21;
{if(i%3==0&
i%7==0)s+=i;
returnsqrt(s);
18、inti;
{a=a/i;
19、inti;
for(i=1,*n=0;
=x;
i+=2)
{if(x%i==0)
{pp[*n]=i;
(*n)++;
20
*c=a/10*100+a%10+b/10*10+b%10*1000;
21、inti;
i++)s+=a->
s[i];
a->
ave=s/N;
22、char*b=a;
while(b<
{if(*b!
{*a=*b;
b++;
while(*p)
{*a=*p;
p++;
23、inti;
i++)s+=a[i].s;
s/=N;
for(i=0,*n=0;
{if(a[i].s<
{b[*n]=a[i];
24、inti;
8;
i++)s+=sqrt((x[i]+x[i+1])/2);
25、inti;
doubles=0,x1=0;
10;
i++)x1+=x[i];
x1/=10;
i++)s+=(x[i]-x1)*(x[i]-x1);
s=sqrt(s/10);
26、inti;
{if(a[i].s>
=s)
27、inti;
for(i=1,*max=a[0],*d=0;
{if(*max<
a[i])
{*max=a[i];
*d=i;
28、inti=1;
while(a>
eps)
{a=a*i/(2*i+1);
returns*2;
29、inti;
{s+=1.0/(i*(i+1));
30、inti,j;
for(i=0,j=0;
{if(a[i].s>
=l&
a[i].s<
=h)
{b[j]=a[i];
returnj;
31、inti=0,j=0;
{if(a[i]=='
)j++;
elsej=0;
if(j>
{a[i-j+n]='
32、
/*以下代码仅供参考*/
inti,len;
/*len为串长*/
longx=0;
len=strlen(p);
/*以下完成数字字符串转换为一个数字。
注意:
字符'
不是数字0*/
len;
i++)x=x*10+p[i]-'
returnx;
33、
inti=0,k;
k=i;
while(a[i]!
)/*以下程序段实现非*字符前移*/
{if(a[i]!
{a[k]=a[i];
k++;
a[k]='
34、inti;
*k=0;
t;
if(s[i]>
s[*k])*k=i;
35、inti=0,j=0;
{if(s[i]%2==0)
36、inti=0,j=0;
)j=0;
elsej++;
a[i-j]=0;
37、/*以下代码仅供参考*/
intj=0,t=m+1;
while(j<
k)
{
/*按题目要求完成以下代码*/
if(prime(t))
{xx[j]=t;
t++;
38、inti,s=a[0][0];
M;
if(a[0][i]>
s)s=a[0][i];
if(a[1][i]>
s)s=a[1][i];
39、inti;
STRECt={'
-1};
{if(strcmp(a[i].num,b)==0)t=a[i];
returnt;
40、/*以下代码仅供参考*/
char*p,*q;
intn=0;
p