国家计算机二级C语言机试改错试题及答案

(1)double fun(int n)

(2)s=s+(double)a/b;

(1)for(i=0;str[i];i++)

(2)if(substr[k+1]== '\0')

(1)t = *x ; *x = y ;

(2)return(t) ;或return t;

(1)t=a;a=b;b=t;

(2)return(b);或return b;

(1)*t=0;

(2)if(d%2!=0)或if(d%2==1)

(1)去掉分号

(2)case 1:case 2:return 1;

(1)b[k]=*p;

(2)b[k]=' ';k++;

(1)if(( tt[i]>='a')&&( tt[i] <= 'z'))

(2)tt[i]-=32;

(1)for (j = i + 1; j < 6; j++)

(2)*(pstr + i) = *(pstr + j) ;

(1)long k=1;

(2)num /=10;

(1)float k;

(2)if(*a< *c)

(1)int k=0;

(2)while (*p||*q)

(1)long s=0, t=0;

(2)t=t/10;

(1)a2=k/10;

(2)return i;

(1)void fun(char *a)

(2)printf("%c" ,*a) ;

(1)n= *p-'0';

(2)n=n*8+*p-'0';

(1)if(d%2==0)

(2)s/=10;

(1)while((high>=2) && (n<10))

(2)yes=0; break;

(1)s[j++]=s[i];或{s[j]=s[i];j++;}

(2)s[j]= '\0';

(1)void fun(long s,long *t)

(2)sl=sl*10;

(1)double f(double x)

(2)return s ;

(1)fun(int x,int y, int z)或int fun(int x,int y,int z)

(2)return j;

(1)while( *r)

(2)*a=*r;a++;r++;

(1)char *fun(char (*sq)[M])

(2)return sp；

(1)int fun(int a[],int m)或fun(int a[],int m)

(2)else if(m>a[mid])

(1)void fun(long s，long *t)

(2)while(s>0)

(1)double fun(double a,double x0)

(2)if(fabs(x1-x0)>=0.00001)

(1)p=j ;

(2)p=i;

(1)#define FU(m,n) (m)/(n)

(2)return (value);

(1)n=strlen( aa ) ;

(2)ch=aa[i];

(1)k++;

(2)if(m==k)

(1)void fun ( char *s , char *t )

(2)t[2*d]='\0';或t[d+i]='\0';或t[2*d]=0;或t[d+i]=0;

(1)for(i=2;i<=m;i++)

(2)y+=1.0/(i*i);或y+=(double)1.0/(i*i)

(1)if(i%2||s[i]%2==0) 或 if(i%2 !=0 || s[i]%2==0)

(2)t[j]='\0'; 或 t[j]=0;

(1)if(p==n) return -1;

(2)a[i]=a[i+1];

(1)r=t;

(2)应填：r=t;或r=&t[0];

(1)while(fabs(t)>=num)

(2)t=s/n;

(1)while(*s)或while(*s!='\0')

(2)s++;

(1)s[j++]=s[i];

(2)s[j]= '\0';

(1)double sum=0.0;

(2)sum/=c;

(1)while (*w)

(2)if ( *r ==*p )

(1)t=calloc(m,sizeof(STU));

(2)t[k]=b[j];

(1)double r ;

(2)while (fabs(n-m) >0.001)

(1)int i, sl;

(2)t[i]=s[sl-i-1];

(1)t=1.0;

(2)return(s*2);

(1)for(i=j+1;i

(2)p=i;

(1)if( n==0)

(2)result *=n--;或{result *=n; n--;}

(1)s=s +(double)(n+1)/n;

(2)return t;

(1)sum=0.0;

(2)if((i+1)%5==0)

(1)for(i=0,t=0;p[i];i++)

(2)c[t]=