POJ 2013 Symmetric Order

最新推荐文章于 2020-11-01 14:23:35 发布

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最新推荐文章于 2020-11-01 14:23:35 发布

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文章标签： ACM POJ

本文链接：https://blog.youkuaiyun.com/qq_24653023/article/details/47686833

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本文介绍了一个简单的对称排序问题，该问题要求将按非递减长度顺序排列的名字列表重新组织，使较短的名字位于顶部和底部，较长的名字位于中间。通过对输入数据进行特定处理，实现了这种对称布局。

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Symmetric Order

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 42 Accepted Submission(s) : 12

Problem Description

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.

Example input:	Example output:
7 Bo Pat Jean Kevin Claude William Marybeth 6 Jim Ben Zoe Joey Frederick Annabelle 5 John Bill Fran Stan Cece 0	SET 1 Bo Jean Claude Marybeth William Kevin Pat SET 2 Jim Zoe Frederick Annabelle Joey Ben SET 3 John Fran Cece Stan Bill

Source

Mid-Central USA 2004

注意：

1：就只是一水题

#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]);
            }
        }
    }
}
/*
7
Bo
Pat
Jean
Kevin
Claude
William
Marybeth
6
Jim
Ben
Zoe
Joey
Frederick
Annabelle
5
John
Bill
Fran
Stan
Cece
*/