2014
02-17

There are N+1 points on the plane, number A,1,2…N. N is an odd integer. Initially, there exist (N+1)/2 edges, as shown in the picture below.

Now your mission is add some edges to the graph, makes that
① degree(i) != degree(j), (i != j, 1 <= i, j <= N).
② degree(1) as small as possible.
For example, when N = 3, there are two possible answers:

Each test case contains a single odd integer N(N<=10^6), indicating the number of points. The input is terminated by a set starting with N = 0.

Each test case contains a single odd integer N(N<=10^6), indicating the number of points. The input is terminated by a set starting with N = 0.

3
0

2

#include<stdio.h>
int main()
{
int n;
while(scanf("%d",&n),n)
printf("%d\n",n/2+1);
}

1. 第2题，TCP不支持多播，多播和广播仅应用于UDP。所以B选项是不对的。第2题，TCP不支持多播，多播和广播仅应用于UDP。所以B选项是不对的。

2. 可以根据二叉排序树的定义进行严格的排序树创建和后序遍历操作。如果形成的排序树相同，其树的前、中、后序遍历是相同的，但在此处不能使用中序遍历，因为，中序遍历的结果就是排序的结果。经在九度测试，运行时间90ms，比楼主的要快。

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