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].
Examples
Input
6 3 2 4
Output
5
Input
6 3 1 3
Output
1
Input
5 2 1 5
Output
0
Note
In the first test Luba can do the following operations: shift the mouse cursor to the tab 2, close all the tabs to the left of it, shift the mouse cursor to the tab 3, then to the tab 4, and then close all the tabs to the right of it.
In the second test she only needs to close all the tabs to the right of the current position of the cursor.
In the third test Luba doesn’t need to do anything.
分析：
理解题意错误表示很伤。本题，十分简单。我的代码的基本思路是将pos点的位置分情况考虑。在所给区域之间，左边，右边。尤其是之间的时候要特别注意：pos是先往左边移动还是先往右边移动。很明显先往距离小的那边移动才是最明确的选择。所以需要考虑两种情况。
#include"stdio.h"
#include"string.h"
#include"math.h"
int main()
{
int n,pos,l,r;
int i,j,k,count;
while(~scanf("%d%d%d%d",&n,&pos,&l,&r))
{
count=0;
if(pos>=l&&pos<=r)
{
if(pos-l<r-pos)
{
if(l>1)
{
count=count+pos-l;
count++;
if(r<n)
{
count=count+r-l;
count++;
}
}
else
{
if(r<n)
{
count=count+r-pos;
count++;
}
}
}
else
{
if(r<n)
{
count=count+r-pos;
count++;
if(l>1)
{
count=count+r-l;
count++;
}
}
else
if(l>1)
{
count=count+pos-l;
count++;
}
}
}
else
if(pos>r)
{
count=count+pos-r;
count++;
if(l>1)
{count=count+r-l;
count++;
}
}
else
if(pos<l)
{
count=count+abs(l-pos);
count++;
if(r<n)
{
count=count+r-l;
count++;
}
}
printf("%d\n",count);
}
}
我的代码负责且智障。在我百度了一下大神的代码发现极其简单。`
#include <stdio.h>
#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);
}
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作者：Int32ToByte
来源：CSDN
原文：https://blog.csdn.net/finalcsdn/article/details/79054721
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