C程序设计语言第二版新版第三章自用

女爷i 2021-12-22 09:25 228阅读 0赞

3-1  
\#include<stdio.h>  
int main()  
\{  
return 0;  
\}  
int binsearch(int x,int v\[\],int n)  
\{  
int low,heigh,mid;  
low=0;  
heigh=n-1;  
mid=(low+heigh)/2;  
while(low<=heigh &&x!=v\[mid\])  
\{  
if(x<v\[mid\]) heigh=mid-1;  
else low = mid+1;  
mid=(low+heigh)/2;  
\}  
if(x==v\[mid\])  
return mid;  
else  
return -1;  
\}

3-2  
\#include<stdio.h>  
int main()  
\{  
return 0;  
\}  
void escape(char s\[\],char t\[\])  
\{  
int i,j;  
for(i=j=0;t\[i\]!='\\0';i++)  
\{  
switch(t\[i\])\{  
case '\\n':  
\{s\[j++\]='\\\\';s\[j++\]='n';break;\}  
case '\\t':  
\{s\[j++\]='\\\\';s\[j++\]='t';break;\}  
default:  
\{s\[j++\]=t\[i\];break;\}  
\}  
\}  
s\[j\]='\\0';

\}  
void unsecape(char s\[\],char t\[\])  
\{  
int i,j;  
for(i=j=0;t\[i\]!='\\0';i++)  
\{  
switch(t\[i\])\{  
case '\\\\':  
switch(t\[++i\])\{  
case 'n':  
s\[j++\]='\\n';break;  
case 't':  
s\[j++\]='\\t';break;  
default:  
s\[j++\]='\\\\';s\[j++\]=t\[i\];break;  
\}  
default:  
s\[j++\]=t\[i\];break;  
\}  
\}  
s\[j\]='\\0';  
\}  
3-3  
\#include<stdio.h>  
int main()  
\{  
printf("yes");  
return 0;  
\}  
void expand(char s1\[\],char s2\[\])  
\{  
char c;  
int i,j;  
i=j=0;  
while((c=s1\[i++\])!='\\0')  
\{  
if(s1\[i\]=='-'&& s1\[i+1\]>c)  
\{  
i++;  
while(c<s1\[i\])  
s2\[j++\]=c++;  
\}  
else s2\[j++\]=c;  
\}  
s2\[j\]='\\0';  
\}  
3-4  
\#include<stdio.h>  
\#define abx(x) ((x)<0? -(x):(x))  
int main()  
\{  
printf("yes");  
return 0;  
\}  
void itoa(int n,char s\[\])  
\{  
int i,sign;  
sign=n;  
i=0;  
do\{  
s\[i++\]=abx(n%10) +'0';  
\}while((n/=10)!=0);  
if(sign <0)  
s\[i++\]='-';  
s\[i\]='\\0';  
//reverse(s);  
\}  
3-5  
\#include<stdio.h>  
int main()  
\{  
printf("yes");  
return 0;  
\}  
void itoa(int n,char s\[\],int b)  
\{  
int i,j,sign;  
if((sign=n)<0) n=-n;  
i=0;  
do  
\{  
j=n%b;  
s\[i++\]=(j<=9)? j+'0':j+'a'-10;  
\}while((n/=b)>0);  
if(sign<0) s\[i++\]='-';  
s\[i\]='\\0';  
//reverse(s);  
\}  
3-6  
\#include<stdio.h>  
\#define abx(x) ((x)<0? -(x):(x))  
int main()  
\{  
printf("yes");  
return 0;  
\}  
void itoa(int n,char s\[\],int w)  
\{  
int i,sign;  
sign=n;  
i=0;  
do\{  
s\[i++\]=abx(n%10) +'0';  
\}while((n/=10)!=0);  
if(sign <0)  
s\[i++\]='-';  
while(w>=i)  
s\[i++\]=' ';  
s\[i\]='\\0';  
//reverse(s);  
\}

转载于:https://www.cnblogs.com/changgf/p/10872893.html