C语言模拟题库改1Word格式.docx
- 文档编号:6123476
- 上传时间:2023-05-06
- 格式:DOCX
- 页数:48
- 大小:832.87KB
C语言模拟题库改1Word格式.docx
《C语言模拟题库改1Word格式.docx》由会员分享,可在线阅读,更多相关《C语言模拟题库改1Word格式.docx(48页珍藏版)》请在冰点文库上搜索。
=0)
m1=n1;
n1=a;
最大公约数是%d\n"
n1);
最小公倍数%d\n"
m*n/n1);
程序改错题:
#include<
main()
{
floata[4][4],t;
inti,j;
printf("
Input4*4numbers:
for(i=0;
i<
4;
for(j=0;
j<
j++)
/************found************/
%d"
a[i][j]);
t=a[i][i];
a[i][j]*=temp;
%.2f"
a[i][j]);
答案:
(1)scanf("
%f"
&
a[i][j]);
(2)a[i][j]/=t;
#include<
stdlib.h>
math.h>
main()
{inti,j,k;
for(i=0;
i<
=2;
i++)
/************found************/
j<
=5;
for(k=0;
k<
=10;
k++)
/************found************/
{if(i*5+j*2+k*1=10)
%d,%d,%d\n"
i,j,k);
(1)for(j=0;
j++)
(2){if(i*5+j*2+k*1==10)
#include<
string.h>
charstr[256];
inti,k=0,n;
gets(str);
n=strlen(str);
n;
/************found************/
if(str[i]>
=0&
&
str[i]<
=9)
{str[k]=str[i];
n++;
str[k]='
\0'
;
printf("
%s\n"
str);
(1)if(str[i]>
=’0’&
=’9’)
(2)k++;
voidfun(char*s,char*t)
inti,s1=strlen(s);
s1;
t[i]=s[i];
t[s1+i]=t[s1-i-1];
t[s1+i]="
\0"
charstr1[40],str2[80];
%c"
str1[40]);
fun(str1,str2);
puts(str2);
(1)t[s1+i]=’\0’;
(2)gets(str1);
或者scanf("
%s"
str1);
{inti,j,t;
for(i=1;
=500;
{t=0;
for(j=1;
if(i%j==0)t+=j;
if(t==i)
{printf("
\n%disafullnumber\n"
i);
factorsof%dare:
\t"
j;
i++)
if(i%j==0)printf("
%d,"
j);
(1)for(j=1;
i;
(2)for(j=1;
{inta[3][3],i,j,max;
3;
a[3][3]);
Themaxis:
"
for(j=0;
max=a[0][0];
for(i=0;
if(max<
a[i][j])max=a[i][j];
%3d"
max);
(2)max=a[0][j];
{intscore,t;
Pleaseenterascore:
/*************found**************/
score);
while(score<
0||score>
100);
t=score/10;
switch(t)
{case10:
case9:
Excellent!
break;
case8:
Good!
case7:
Middle!
case6:
Pass!
defaut:
Fail!
\n);
答案:
(1)scanf("
(2)default:
\n”);
{inta[10]={16,1,2,7,3,12,15,34,5,11},i,s;
s=0;
10;
if(i%2==1)
s=s+a[i];
print("
Theresultis:
%.2f\n"
s);
(1)if(a[i]%2==1)
(2)printf("
%d\n"
#include"
stdio.h"
intx,max,min,i;
=9;
x);
if(i=1){max=x;
min=x;
if(x>
max)max=x;
max)min=x;
max=%d,min=%d\n"
max,min);
(1)if(i==1){max=x;
(2)if(min>
x)min=x;
程序填空题:
intn,m=0,s,c;
scanf("
while(m<
20)
s=n;
c=0;
while(s!
=0)
c=c+s%10;
s=s/10;
/**************found************/
___
(1)___
{m++;
%5d"
n);
___
(2)___
(1)if(c==5);
(2)n++;
{inti,j;
=5-i;
____
(1)____
=i;
'
A'
+j-1);
for(j=i-1;
j>
=1;
j--)
____
(2)____
(1)printf(“”);
{inti,j,p,x[][4]={1,5,7,4,2,6,4,3,8,2,3,1};
{
___
(1)___
if(___
(2)___)
p=j;
Theminvalueinline%dis%d\n"
i,x[i][p]);
(1)p=0;
(2)x[i][p]>
x[i][j]
intn,p,a;
inputn:
/**********found*********/
___
(1)___
while(n!
a=___
(2)___;
if(a%2!
=0)p+=a;
n=n/10;
p);
(2)n%10;
inti,j;
charch;
=4;
putchar('
'
/**************found************/
___
(1)___;
for(j=9-2*i;
0;
___
(2)___);
\n'
(1)ch=’b’;
(2)ch++
main()
inti,a[10],t;
/**************found************/
___
(2)___
t=a[i];
a[i]=a[9-i];
a[9-i]=t;
a[i]);
(1)scanf(“%d”,&
a[i]);
(2)for(i=0;
5;
conio.h>
intprinum(intb[10])
{ints,i,j,k;
k=b[i]-1;
for(j=2;
=k;
if(____
(1)____)
break;
if(j<
=k)
returns;
{inta[]={2,3,4,5,6,7,9,10,11,13},n;
n=prinum(a);
非素数之积为:
%d\n"
n);
(1)b[i]%j==0
(2)s=s+b[i];
inta,b,t,k;
Pleaseentertwonumbers:
a,&
b);
if(a<
b)
{t=a;
a=b;
b=t;
k=a%b;
while(____
(1)____)
{a=b;
b=k;
GCDis:
b);
(1)k
(2)k=a%b;
{inti,p=0,a[10];
___
(1)___;
p=i;
%d,%d\n"
a[p],p);
(1)i<
10
(2)a[p]>
a[i]
程序设计题:
voidnono(ints);
intfun(inta[3][3])
inti,j,s;
/***********begin***********/
/************end************/
{inta[3][3]={1,2,3,4,9,5,7,8,6};
inti,j,s=0;
arrayis:
s=fun(a);
Resultis:
s);
nono(s);
voidnono(ints)
{FILE*f;
f=fopen("
out1.dat"
"
w"
fprintf(f,"
Resultis:
fclose(f);
s=0;
3;
for(j=0;
j<
{
if(i==j)s=s+a[i][j];
elseif(i+j==2)s=s+a[i][j];
}
#include"
math.h"
conio.h"
stdlib.h"
voidNONO(doublex);
{doubles;
inti,n;
s=%f\n"
NONO(s);
voidNONO(doublex)
{FILE*f;
x);
for(i=1;
=n;
s=s+1.0/(i*i);
voidNONO(intmax);
{inta,b,c,max;
max=%d\n"
max);
NONO(max);
voidNONO(intmax)
%d,%d,%d"
a,&
b,&
c);
if(a>
b)max=a;
elsemax=b;
if(max<
c)max=c;
voidnono(floats);
floata,b,c,area,x;
/***********begin***********/
/************end************/
nono(area);
voidnono(floats)
%.2f\n"
area=0;
%f,%f,%f"
x=(a+b+c)/2;
if(a+b>
c)
if(a+c>
b)
if(b+c>
a)area=sqrt(x*(x-a)*(x-b)*(x-c));
if(area)printf("
%.2f"
area);
elseprintf("
No"
voidNONO(longs);
voidmain()
intn,i,j;
longp,s;
intf=1;
s=%ld\n"
voidNONO(longs)
FILE*wf;
wf=fopen("
);
fprintf(wf,"
s);
fclose(wf);
p=1;
=2*n-1;
i=i+2)
for(j=1;
=i;
p=p*j;
s=s+f*p;
p=1;
f=-f;
voidNONO(intcount);
{i
- 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:
- 语言 模拟 题库