3计算机国二题库缩减版Word文档格式.docx
- 文档编号:17893665
- 上传时间:2022-12-11
- 格式:DOCX
- 页数:25
- 大小:19.92KB
3计算机国二题库缩减版Word文档格式.docx
《3计算机国二题库缩减版Word文档格式.docx》由会员分享,可在线阅读,更多相关《3计算机国二题库缩减版Word文档格式.docx(25页珍藏版)》请在冰豆网上搜索。
(1)(fp)
(2)==(3)fp
13/*********found**********/
for(i=0;
i<
__1__;
i++)
(1)N
(2)substr(3)0
14/*********found**********/
for(p=1;
p<
=__1__;
p++)
(1)k
(2)N-1(3)temp
15/*********found**********/
t=(a>
b)?
(b>
c?
b:
(a>
c:
___1___)):
((a>
c)?
___2___:
((b>
___3___));
(1)a
(2)a(3)b
16/*********found**********/
if((ch>
a'
)___1___(ch<
z'
))
(1)&
&
(2)'
A'
(3)ch
17/*********found**********/
if(isdigit(*s))sum+=*s-__1__;
(1)48
(2)s++(3)sum
18/*********found**********/
f=___1___;
(1)1
(2)-1(3)t
19/*********found**********/
if(n>
=len)strcpy(__1__);
(1)t,s
(2)s[i](3)0
20/*********found**********/
__1__fun(structstudent*a)
(1)structstudent*
(2)a->
score[i](3)a
21/*********found**********/
max=min=___1___;
(1)a[i]
(2)a[j](3)a[j]
22/*********found**********/
voidfun(int___1___,intn)
(1)t[][N]
(2)i=0;
i<
n(3)s
23/*********found**********/
voidfun(int___1___,intn)
(1)*a
(2)2(3)i+1
24/*********found**********/
f=1.0+___1___;
(1)x
(2)n(3)t
25/*********found**********/
(同第1题一样)
for(i=___2___;
n;
(1)1
(2)1(3)i
26/*********found**********/
b__1__=10004;
(1)->
sno
(2)->
name(3)&
t
27/*********found**********/
k=__1__;
(同第49题一样)
/*********found**********/
t=__2__;
(1)1
(2)2*i(3)(-1)
28/*********found**********/
s=__1__;
(1)0
(2)n(3)(t*t)
29/*********found**********/
ch=tolower(__1__);
(1)*s
(2)1(3)k[n]
30/*********found**********/
__1__t;
(1)structstudent
(2)n-1
(3)a[i].name,a[j].name
31/*********found**********/
if(___1___==0){
(1)a[i]%2
(2)a[j](3)j
32/*********found**********/
{t=*n%__1__;
(1)10
(2)0(3)x
33/*********found**********/
i++)(同第36和第39一样)
if(len<
=__2__)
(1)N
(2)k(3)ss[i]
34/*********found**********/
flag=1;
}
(1)n++
(2)0(3)s++
35/*********found**********/
t[j]=__1__;
j++;
(1)s[i]
(2)k(3)0
36/*********found**********/
for(i=0;
if(strcmp(ss[i],t)==0)
return__2__;
(同第33和第39一样)
(1)N
(2)i(3)-1
37/*********found**********/
lp=__1__;
(1)s
(2)--(3)return0
38/*********found**********/
x=__1__/4;
(1)3.
(2)>
(3)(2*i+1)
39/*********found**********/
for(i=0;
___1___;
a[i]=a[n-1-___2___];
(同第33和第36一样)
(1)n/2
(2)i(3)a[n-i-1]
40/*********found**********/
b=__1__;
(1)a
(2)b.name(3)score[i]
41/*********found**********/
for(___1___;
j<
i;
j++)
(1)j=2
(2)i(3)j
42/*********found**********/
(1)1
(2)s(3)i*10
43/*********found**********/
__1__fun(structstudenta)
(1)structstudent
(2)a.name
(3)a.score[i]
44/*********found**********/
p=(n%2==0)?
n/2:
n/2+___1___;
(1)1
(2)i(3)a[p+i]
45/*********found**********/
if(a[i]%2==___1___)
(1)1
(2)j++(3)j
46/*********found**********/
n=__1__;
(1)0
(2)x(3)t++
47/*********found**********/
*(t+n)=__1__;
n++;
(1)*s
(2)s++(3)n
48/*********found**********/
#defineOK(i,t,n)
((___1___%t==0)&
(i/t<
n))
(1)i
(2)t++(3)count
49/*********found**********/
k=__1__;
(同第27题一样)
if(strlen(ps[k])<
strlen(__2__))k=j;
(1)i
(2)ps[j](3)tp
50/*********found**********/
for(j=a[i]*2;
=n;
j+=___1___)
(1)a[i]
(2)a[i](3)0
二、改错题
A1
(1)for(i=1;
=y;
i++)
(2)t=t%1000;
A2
(1)if(t==0)
(2)*zero=count;
A3
(1)*t=0;
(2)if(d%2!
=0)
A4
(1)for(i=2;
=m;
i++)
(2)y-=1.0/i/i;
A5
(1)t=1;
(2)return(2*s);
A6
(1)#defineFU(m,n)((m)/(n))
(2)return(value);
(注:
R,V小写)
A7
(1)for(i=strlen(t)-1;
i;
i--)
(2)if(t[j]>
t[j+1])
A8
(1)voidfun(longs,long*t)
(2)while(s>
0)
A9
(1)doubler;
(2)while(fabs(m-n)>
0.001)(注:
n,m,<
)
A10
(1)b[k]=*p;
(2)b[k]='
'
;
k++;
单引号)
A11
(1)voidfun(char*s,char*t)
(2)t[2*d]=0;
A12
(1)n=strlen(aa);
(2)ch=aa[i];
A13
(1)fun(intx,inty,intz)
(2)returnj;
A14
(1)voidfun(inta[][M],intm)
A
(2)a[j][k]=(k+1)*(j+1);
A15
(1)c=tolower(c)(注:
c小写);
(2)c=c+5;
A16
(1)voidfun(int*a)
(2)a[j]=a[j-1];
A17
(1)if((k%13==0)||(k%17==0))(注:
双等号)
(2)}(注:
横线去掉)
A18
(1)for(i=1;
=3;
i++)
(2)if(k>
=0&
k<
=6)
A19
(1)if(i%k==0)
=i)
A20
(1)char*fun(char(*sq)[M])
(2)returnsp;
A21
(1)switch(g)
(2)case1:
return1;
case2:
return1;
A22
(1)fun(inta[],intm)
(2)elseif(m>
a[mid])
A23
(1)sum=0;
(2)if((i+1)%5==0)
A24
(1)d=1;
(2)d=d/10;
A25
(1)k=i;
(2)c=k;
A26
(1)doublefun(doublea,doublex0)
(2)if(fabs(x1-x0)>
0.00001)
A27
(1)k++;
(2)if(m==k)
A28
(1)for(i=0;
str[i];
(2)if(substr[k+1]==0)
A29
(1)floatfun(intk)
(2)returns;
A30
(1)q=p+i;
(2)while(q>
p)
A31
(1)fun(intn)
(2)if(n==1)
A32
(1)if(n==0)
(2)result*=n--;
A33
(1)intk=0;
(2)while(*p||*q)
A34
(1)t+=s[k];
(2)*aver=ave;
A35
(1)while(*w)
(2)if(*r==*p)
A36
(1)a2=k/10;
(2)returni;
A37
(1)doublefun(intn)
(2)c=a;
a+=b;
b=c;
A38
(1)doublefun(intn)
(2)s=s+(double)a/b;
A39
(1)n=*p-'
(2)n=n*8+*p-'
A40
(1)s[j++]=s[i];
(2)s[j]=0;
A41
(1)sum=0;
j=0;
(2)if(sum%4==2)
A42
(1)intfun(intn,intxx[][M])
(2)printf("
%d"
xx[i][j]);
A43
(1)if(p==n)return-1;
(2)a[i]=a[i+1];
A44
(1)floatk;
(2)if(*c>
*a)
A45
(1)t=a;
a=b;
b=t;
(2)return(b);
A46
(1)inti,sl;
(2)t[i]=s[sl-i-1];
A47
(1)intj,c=0;
doublexa=0.;
(2)if(x[j]>
=xa)
A48
(1)intfun(inta,intb,intc)
(2)elsereturn1;
A49
(1)doublefun(intn)
(2)returnsum;
A50
(1)y=1;
(2)d=a-i;
三、编程题
1、doublefun(doublex)
{inti=0;
doubles=1,a=1;
while(fabs(a)>
0.000001)
{a=a*(0.5-i)*x/(i+1);
s+=a;
i++;
returns;
2、doublefun(intn)
{inti;
doubles=0,a=1;
for(i=1;
{a=a/i;
3、voidfun(charp1[],charp2[])
{while(*p1)p1++;
while(*p2)
{*p1=*p2;
p1++;
p2++;
*p1=0;
4、intfun(intscore[],intm,intbelow[])
{inti=0,s=0,n=0;
m;
i++)s+=score[i];
s/=m;
{if(score[i]<
s)
{below[n]=score[i];
returnn;
5、voidfun(char*a,intn)
{
/*以下代码仅供参考*/
inti=0,j,k=0;
while(a[k]=='
*'
)k++;
/*k为统计*字符个数*/
if(k>
n)
{i=n;
j=k;
/*以下完成将下标为k至串尾的字符前移k-n个位置*/
while(a[j])
{a[i]=a[j];
a[i]=0;
6、voidfun(char*a)
{inti=0,j=0;
while(a[i]=='
)i++;
while(a[i])
{a[j]=a[i];
a[j]=0;
7、voidfun(char*ss)
{inti=1;
while(ss[i-1]&
ss[i])
{if(ss[i]>
ss[i]<
)ss[i]+='
-'
i+=2;
8、voidfun(STRECa[])
{STRECtmp;
inti,j;
for(i=0;
i<
N;
for(j=i+1;
j<
{/*请按题目要求完成以下代码*/
if(a[i].s<
a[j].s)
{tmp=a[i];
a[i]=a[j];
a[j]=tmp;
9、voidfun(char*a,char*h,char*p)
{a=h;
while(h!
=p)
{if(*h!
{*a=*h;
a++;
h++;
while(*p)
{*a=*p;
p++;
*a=0;
10、doublefun(STREC*h)
doubles=0;
while(h->
next)
{h=h->
next;
s+=h->
s;
s/=i;
11、voidfun(char*s,chart[])
while(s[i])
{if(i%2==1&
s[i]%2==1)
{t[j]=s[i];
t[j]=0;
12、doublefun(STREC*h)
{doubles=0;
if(s<
h->
s)s=h->
13、voidfun(char*a,intn,inth,inte)
n-h-e;
i++)a[i]=a[i+h];
14、voidfun(inta,intb,long*c)
{*c=a/10*10+a%10*1000+b/10+b%10*100;
15、doublefun(intn)
doubles=0,a=0;
{a+=sqrt(i);
16、intfun(intn)
{inti,s=0;
for(i=2;
{if(n%i==0)s+=i;
17、doublefun(intn)
for(i=21;
{if(i%3==0&
i%7==0)s+=i;
returnsqrt(s);
18、doublefun(intn)
19、voidfun(intx,intpp[],int*n)
for(i=1,*n=0;
=x;
i+=2)
{if(x%i==0)
{pp[*n]=i;
(*n)++;
20、voidfun(inta,intb,long*c)
{*c=a/10*100+a%10+b/10*10+b%10*1000;
21、voidfun(STREC*a)
N;
i++)s+=a->
s[i];
a->
ave=s/N;
22、voidfun(char*a,char*p)
{char*b=a;
while(b<
{if(*b!
{*a=*b;
b++;
{*a=*p;
p++;
23、doublefun(STREC*a,STREC*b,int*n)
i++)s+=a[i].s;
s/=N;
for(i=0,*n=0;
{if(a[i].s<
{b[*n]=a[i];
24、doublefun(doublex[9])
8;
i++)s+=sqrt((x[i]+x[i+1])/2);
25、doublefun(doublex[10])
doubles=0,x1=0;
10;
i++)x1+=x[i];
x1/=10;
i++)s+=(x[i]-x1)*(x[i]-x1);
s=sqrt(s/10);
26、doublefun(STREC*a,STREC*b,int*n)
{if(a[i].s>
=s)
- 配套讲稿:
如PPT文件的首页显示word图标,表示该PPT已包含配套word讲稿。双击word图标可打开word文档。
- 特殊限制:
部分文档作品中含有的国旗、国徽等图片,仅作为作品整体效果示例展示,禁止商用。设计者仅对作品中独创性部分享有著作权。
- 关 键 词:
- 计算机 题库 缩减