Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 7163    Accepted Submission(s): 2772

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

 

Source

 

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题目大意：数字依照所给图形进行排列。给出两点。求两点距离。

解题思路：图画大一点，然后找规律：固定这个数字的三维坐标。行非常好找。直接将这个数开根号就能够知道是第几行，假设开根号正好是整数的话就直接是行号，否则须要加1，对于列有两个方向都须要进行求解。详细例如以下图：

第一种右斜着的列号：

watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQv/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/gravity/Center" alt="">另外一种是左斜着的列号：

依据这两种进行找规律。

详见代码。

#include <iostream>

#include <cstdio>

#include <cmath>

using namespace std;

int abs(int a)

{

    if (a>0)

        return a;

    else

        return -a;

}

int main()

{

    int a,b;

    while (~scanf("%d%d",&a,&b))

    {

        int h1=sqrt(a);

        if (h1*h1==a)

            h1=h1;

        else

            h1=h1+1;

        int h2=sqrt(b);

        if (h2*h2==b)

            h2=h2;

        else

            h2=h2+1;

        int r1=h1*h1;

        int rx1=(r1-a)/2+1;

        int r2=h2*h2;

        int rx2=(r2-b)/2+1;

        int l1=r1-(2*h1-1)+1;

        int lx1=(a-l1)/2+1;

        int l2=r2-(2*h2-1)+1;

        int lx2=(b-l2)/2+1;

        int ans=abs(h1-h2)+abs(lx1-lx2)+abs(rx1-rx2);

        cout<<ans<<endl;

    }

    return 0;

}