HDu 4572 Bottles Arrangement(法則を探す)

Bottles Arrangement
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 224    Accepted Submission(s): 176
Problem Description
Hunan cuisine is really wonderful! But if you don’t like spicy food, you will feel terrible since it can be hard for you to find any food without hot pepper here. Big Fan is a student from the north who was not fit to the spicy food in Changsha. He became thinner and thinner because eating little food and maintained his life mostly by drinking water. One day, he found that the wine in Hunan is pretty good, such as Jiugui, Liuyang River, Shaoyang Daqu and so on. He got addicted to it and became an alcoholic, leading a depressed life. 
Now N days have passed and he is sobered. He is surprised to find that there are exactly N×M bottles around him. Another amazing fact is that there are N bottles with height 1 and N bottles with height 2 … N bottles with height M.
Now he is interested in playing with these bottles. He wants to arrange all these bottles in a rectangle with M rows and N columns which satisfied: 
(1)In any column, there are no bottles with same height; 
(2)In any row, the height difference between any two adjacent bottles is no more than 1. 
He defined a strange function Y which equals the maximum value of the total height of any single row. He is addicted in arranging these rubbish bottles to find the minimal Y. You know that he cannot solve it with his pour IQ. You are his friend and can’t endure his decadence any more. So you decide to help him solve this problem and then bring him back to study.
 
 
Input
There are several test cases. For each case, the input contains one line with two integers M and N (1< M <= 10000, 3 <= N < 2×M, It is guaranteed that N is always odd).
The input will finish with the end of file.
 
 
Output
For each test case, print the minimal Y in single line.
 
 
Sample Input
3 3 3 5
 
 
Sample Output
8 11
Hint
For the first case the solution is: 1 2 3 2 1 1 3 3 2
 
 
Source
2013 ACM-ICPC長沙試合区全国招待試合——テーマ再現
 
この问题は简単に书いて、しかし推理は简単ではありませんて、试合の时ただ简単ではないと思ってそれをしていないで、谁がこんなに短いコードが过ごすことができることを知っていて、自分の能力が足りないとしか言えません
 
これは法則を探していたが,その時は直接スキップした.m,m,m-1,m-1,m-2,m-2,m-3,m-3.......前のn項を求めて出力すればよい
この問題の証明について、ネット上で見つけることができて、ある人は見て、証明これは2ページ書いて、どうせこのコードよりずっと長いです
 

#include<stdio.h>

int main()

{

	int n,m,i,sum,j;

	while(scanf("%d%d",&m,&n)!=EOF)

	{

		sum=0;

		for(i=0;i<=n/2;i++)

		{

			sum+=(m-i)*2;

		}

		sum-=m-i+1;

		printf("%d
",sum);

	}

	return 0;

}