codeforce 915 B Browser

Luba is surfing the Internet. She currently has n opened tabs in her browser, indexed from 1 to n from left to right. The mouse cursor is currently located at the pos-th tab. Luba needs to use the tabs with indices from l to r (inclusive) for her studies, and she wants to close all the tabs that don’t belong to this segment as fast as possible.

Each second Luba can either try moving the cursor to the left or to the right (if the cursor is currently at the tab i, then she can move it to the tab max (i − 1, a) or to the tab min (i + 1, b)) or try closing all the tabs to the left or to the right of the cursor (if the cursor is currently at the tab i, she can close all the tabs with indices from segment [a, i − 1] or from segment [i + 1, b]) . In the aforementioned expressions a and b denote the minimum and maximum index of an unclosed tab, respectively. For example, if there were 7 tabs initially and tabs 1, 2 and 7 are closed, then a  = 3, b  = 6.

What is the minimum number of seconds Luba has to spend in order to leave only the tabs with initial indices from l to r inclusive opened?
Input

The only line of input contains four integer numbers n , pos , l , r ( 1   ≤   n   ≤   100 , 1   ≤   pos   ≤   n , 1   ≤   l   ≤   r   ≤   n ) — the number of the tabs, the cursor position and the segment which Luba needs to leave opened.
Output

Print one integer equal to the minimum number of seconds required to close all the tabs outside the segment [l, r] .

题意：浏览器有n个标签页，当前鼠标在第 pos 个标签页上，每一次可以有下面四种操作：

• 移动到左边一个标签页
• 移动到右边一个标签页
• 关掉鼠标所在标签页的所有左边的标签页
• 关掉鼠标所在标签页的所有右边的标签页

每次操作花费都是1。现在想要关掉除了 [l,r] 以外的所有标签页，求最小花费。

#include 
#define min(a,b) ((a)>(b)?(b):(a))
int n, pos,cl,cr;
int main() {
    scanf("%d %d %d %d", &n,&pos,&cl,&cr);
    if (cl == 1 && cr == n)printf("0");
    else if (cl == 1)printf("%d", abs(cr - pos) + 1);
    else if (cr == n)printf("%d", abs(pos - cl) + 1);
    else printf("%d", min(abs(pos - cl), abs(pos - cr)) + cr - cl + 2);
}

题意理解错了很伤啊…唉