HDU 1030 纯数学 找规律

题目如下：大意是求两个三角形之间要通过多少条边

Delta-wave

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4938    Accepted Submission(s): 1875

Problem Description

A triangle field is numbered with successive integers in the way shown on the picture below. 

The traveller needs to go from the cell with number M to the cell with number N. The traveller is able to enter the cell through cell edges only, he can not travel from cell to cell through vertices. The number of edges the traveller passes makes the length of the traveller's route.  Write the program to determine the length of the shortest route connecting cells with numbers N and M. 

Input

Input contains two integer numbers M and N in the range from 1 to 1000000000 separated with space(s).

Output

Output should contain the length of the shortest route.

Sample Input

6 12

Sample Output

3

#include<stdio.h>
#include<math.h>
void main()
{
	int m,n,h1,h2,x,y,temp,num,p1,p2;
	while(scanf("%d %d",&m,&n)!=EOF)
	{
		h1=sqrt((float)m);
		if(h1!=sqrt((float)m))
			h1++;
		h2=sqrt((float)n);
		if(h2!=sqrt((float)n))
			h2++;
		x=m-(h1-1)*(h1-1);
		y=n-(h2-1)*(h2-1);
		p1=(h1*h1-m)/2+1;
		p2=(h2*h2-n)/2+1;
		if(h1>h2)
		{
			temp=h2;
			h2=h1;
			h1=temp;
			temp=x;
			x=y;
			y=temp;
			temp=p1;
			p1=p2;
			p2=temp;
		}
		num=(h2-h1-1)*2;
		if(x%2==0)
			num++;
		if(y%2==0&&x<=y&&p1<=p2)
			num++;
		else
			num+=2;
		if(x>y)
			num+=x-y;
		if(p1>p2)
			num+=y-2*(h2-p1)-1;
		printf("%d\n",num);
	}
}

主要是分两个方向划分到同一个正三角形里面，然后中间的行数*2+头尾吧，头尾三角形分为正反两种，分情况讨论，也不怎么能说清楚啊。。。