xynuoj 对称排序

冷不防 2024-02-18 21:45 13阅读 0赞

### **1971: 对称排序** ###

时间限制: 1 Sec 内存限制: 64 MB  
提交: 5 解决: 2  
您该题的状态：**已完成**  
\[[提交][Link 1]\]\[[状态][Link 2]\]\[[讨论版][Link 3]\]

### **题目描述** ###

In your job at Albatross Circus Management (yes, it's run by a bunch of clowns), you have just finished writing a program whose output is a list of names in nondescending order by length (so that each name is at least as long as the one preceding it). However, your boss does not like the way the output looks, and instead wants the output to appear more symmetric, with the shorter strings at the top and bottom and the longer strings in the middle. His rule is that each pair of names belongs on opposite ends of the list, and the first name in the pair is always in the top part of the list. In the first example set below, Bo and Pat are the first pair, Jean and Kevin the second pair, etc.

### **输入** ###

The input consists of one or more sets of strings, followed by a final line containing only the value 0. Each set starts with a line containing an integer, n, which is the number of strings in the set, followed by n strings, one per line, NOT SORTED. None of the strings contain spaces. There is at least one and no more than 15 strings per set. Each string is at most 25 characters long.

### **输出** ###

For each input set print "SET n" on a line, where n starts at 1, followed by the output set as shown in the sample output. If length of two strings is equal,arrange them as the original order.(HINT: StableSort recommanded)

### **样例输入** ###

7
 Bo
 Pat
 Jean
 Kevin
 Claude
 William
 Marybeth
 6
 Jim
 Ben
 Zoe
 Joey
 Frederick
 Annabelle
 5
 John
 Bill
 Fran
 Stan
 Cece
 0

### **样例输出** ###

SET 1
 Bo
 Jean
 Claude
 Marybeth
 William
 Kevin
 Pat
 SET 2
 Jim
 Zoe
 Frederick
 Annabelle
 Joey
 Ben
 SET 3
 John
 Fran
 Cece
 Stan
 Bill

### **提示** ###

### **来源** ###

[nyoj语言入门][nyoj]

\#include<stdio.h> 
\#include<string.h> 
int main() 
\{ 
int n; 
int i,j,t=1; 
while(~scanf("%d",&n)) 
\{ 
if(n==0) 
break; 
char a\[n\]\[30\]; 
char b\[30\]; 
for(i=0;i<n;i++) 
scanf("%s",a\[i\]); 
for(i=0;i<n-1;i++) 
for(j=0;j<n-1-i;j++) 
if(strlen(a\[j\])>strlen(a\[j+1\])) 
\{ 
strcpy(b,a\[j\]); 
strcpy(a\[j\],a\[j+1\]); 
strcpy(a\[j+1\],b); 
\} 
//for(i=0;i<n;i++) 
//printf("%s\\n",a\[i\]); 
printf("SET %d\\n",t); 
for(i=0;i<n;i+=2) 
printf("%s\\n",a\[i\]); 
if(n%2==0) 
\{ 
for(i=n-1;i>0;i-=2) 
printf("%s\\n",a\[i\]); 
 
\} 
else 
\{ 
for(i=n-2;i>0;i-=2) 
printf("%s\\n",a\[i\]); 
\} 
t++; 
\} 
return 0; 
\}

这道题是我的耻辱，我连冒泡排序都忘了，

结题思路：先排序，在输出，先输出0,2,4,6.....,再判断字符串有几个，若为奇数，则从奇数-2开始倒着输出；否则从个数-1开始输出。

[Link 1]: http://xyoj.xynu.edu.cn/submitpage.php?id=1971
[Link 2]: http://xyoj.xynu.edu.cn/problemstatus.php?id=1971
[Link 3]: http://xyoj.xynu.edu.cn/bbs.php?pid=1971
[nyoj]: http://xyoj.xynu.edu.cn/problemset.php?search=nyoj%E8%AF%AD%E8%A8%80%E5%85%A5%E9%97%A8