Symmetric Order

╰+攻爆jí腚メ 2022-08-18 02:23 54阅读 0赞

## Symmetric Order ##

####  ####

##### Time Limit: 1000ms Memory limit: 65536K 有疑问？点这里^\_^ #####

## 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, 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.

## 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.

## 示例输入 ##

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

## 示例输出 ##

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 ##

面向对数据结构和算法不太了解的同学

## 示例程序 ##

#include<stdio.h>  
    #include<string.h>  
    int main (void)  
    {  
        int n,i,j,leng[40];  
        char a[30][30];  
        int dd=1;  
        while(~scanf("%d",&n)&&n)  
        {  
            for(i=1; i<=n; i++)  
            {  
                scanf("%s",a[i]);  
            }  
            printf("SET %d\n",dd);  
            dd++;  
            for(i=1; i<=n; i++)  
            {  
                if(2*i-1<=n)  
                {  
                    printf("%s\n",a[i*2-1]);  
                }  
            }  
            if(n%2==1)  
            {  
                for(i=1; i<=n/2; i++)  
                {  
                    printf("%s\n",a[n-(i*2-1)]);  
                }  
            }  
            if(n%2==0)  
            {  
                for(i=1;i<n;i+=2)  
                {  
                    printf("%s\n",a[n-i+1]);  
                }  
            }  
        }  
    }