Hduoj2800【水题】

Adding Edges

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 768    Accepted Submission(s): 509

Problem Description

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:

 

Input

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.

 

Output

For each test case, output on a line the smallest possible value of the degree(1).

 

Sample Input

30

 

Sample Output

2

 

Source

HDU 2009-4 Programming Contest 

#include<stdio.h>#include<string.h>int n; int main(){int i;while(scanf("%d", &n) != EOF && n){printf("%d\n", (n+1) >> 1);}return 0;}

题意：给出一对对互相带连线的点，现在要求给他们互相之间连线，要求除了A以外的任意2个点之间都不能都相同的边数，即度数，最后使得编号为1的点的度数最小，求这个最小的度数。

思路：这里总共有n+1个点，除了A以外有n个点，所以对于每个度数不同的点来说必定是从1~n的，那么我们假设某个点i的度数为n，因为它的对面的点已经占了一个度，所以我们必须从剩下的n-1个点与它相连，那么这剩下的n-2个点的度数都变成了2，除了i对面的那个点的度数为1，由于必定会有度数为1的点所以我们不能再动那个点。以此类推，由于1的对面是A，是不计入度数的，所以1能够取得的最小的度数就是n+1的一半。