Symmetric Order C语言 ZOJ2172

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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.

Input

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, sorted in nondescending order by length. 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. 

Output

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.

这道题可以按输入的名字是奇数个还是偶数个分别处理，注意在读入人名的个数后要去掉换行符。

 1 #include <stdio.h>
 2 #include <malloc.h>
 3 #include <string.h>
 4 
 5 int main(void){
 6     int n;//人名数量
 7     int i=1;//SET数
 8     int j;//用于循环
 9     char **name,t;//名字的二维数组的指针
10     while(scanf("%d",&n)==1){
11         scanf("%c",&t);//读入换行符 
12         if(n==0) break;
13         else{
14             name=(char**)malloc(n*sizeof(char*));
15             for(j=0;j<n;j++) name[j]=(char*)malloc(26*sizeof(char));
16             for(j=0;j<n;j++) gets(name[j]);
17             printf("SET %d\n",i++);//SET数加一 
18             //按奇、偶分别输出 
19             if(n%2==0){
20                 for(j=0;j<n;j+=2) puts(name[j]);
21                 for(j=n-1;j>0;j-=2) puts(name[j]);
22             }else{
23                 for(j=0;j<n;j+=2) puts(name[j]);
24                 for(j=n-2;j>0;j-=2) puts(name[j]);
25             }
26         }
27         for(j=0;j<n;j++) free(name[j]);
28         free(name);
29     }
30     return 0;
31 }

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Symmetric Order C语言 ZOJ2172

标签：bsp   one   str   name   ons   oid   sam   rac   desc   

原文地址：https://www.cnblogs.com/20174317zhuyuan/p/9419313.html