Codeforces Beta Round #76 (Div. 1 Only)——A——Frames

hroughout Igor K.'s life he has had many situations worthy of attention. We remember the story with the virus, the story of his mathematical career and of course, his famous programming achievements. However, one does not always adopt new hobbies, one can quit something as well.

This time Igor K. got disappointed in one of his hobbies: editing and voicing videos. Moreover, he got disappointed in it so much, that he decided to destroy his secret archive for good.

Igor K. use Pindows XR operation system which represents files and folders by small icons. At that, m icons can fit in a horizontal row in any window.

Igor K.'s computer contains n folders in the D: disk's root catalog. The folders are numbered from 1 to n in the order from the left to the right and from top to bottom (see the images). At that the folders with secret videos have numbers from a to b inclusive. Igor K. wants to delete them forever, at that making as few frame selections as possible, and then pressing Shift+Delete exactly once. What is the minimum number of times Igor K. will have to select the folder in order to select folders from a to b and only them? Let us note that if some selected folder is selected repeatedly, then it is deselected. Each selection possesses the shape of some rectangle with sides parallel to the screen's borders.

Input

The only line contains four integers n, m, a, b (1 ≤ n, m ≤ 109, 1 ≤ a ≤ b ≤ n). They are the number of folders in Igor K.'s computer, the width of a window and the numbers of the first and the last folders that need to be deleted.

Output

Print a single number: the least possible number of times Igor K. will have to select the folders using frames to select only the folders with numbers from a to b.

Sample test(s)

input

11 4 3 9

output

3

input

20 5 2 20

output

2

Note

The images below illustrate statement tests.

The first test:

In this test we can select folders 3 and 4 with out first selection, folders 5, 6, 7, 8 with our second selection and folder 9 with our third, last selection.

The second test:

In this test we can first select all folders in the first row (2, 3, 4, 5), then — all other ones.

大意：水题，很烦~见大神写法对处理行，列很好

1.m=1时显然次数最小值为1.

2.b=n时特判：如果a在第一列或者a,b同行，次数为1，否则次数为2

3.所有a在第1列，b在最后一列的情况下或a,b同行，最小次数都是1.

4.a在第一列或b在最后一列或者a,b的列相邻(对角的两个矩形)或者a,b所在行相邻则次数为2

5.剩下的次数都是3.

原博文见->戳这里-<

#include <cstdio>

#include <iostream>

using namespace std;

const int maxn =1000+5;

int main()

{

    long long n,m,a,b;

    cin>>n>>m>>a>>b;

    int st=a/m+(a%m!=0);int ed=b/m+(b%m!=0);

    if(m==1){

        cout<<1;

        return 0;

    }

    if(b==n){

        if(a%m==1||st==ed) cout<<1;

        else cout<<2;

        return 0;

    }

    if((a%m==1&&b%m==0)||st==ed) {cout<<1;return 0;}

    if(a%m==1||b%m==0||st+1==ed||(b+1)%m==a%m) cout<<2;

    else cout<<3;

    return 0;

}

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