915B. Browser

xiaoxiao2021-02-28  19

B. Browser time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

Luba is surfing the Internet. She currently has n opened tabs in her browser, indexed from 1 to nfrom 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 12 and 7 are closed, then a = 3b = 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 nposlr (1 ≤ n ≤ 1001 ≤ pos ≤ n1 ≤ 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.

题意:cf 读题真的很恼火,可能我不仅代码能力弱,英语更弱,两个小时只做了AB题,读题都要花一半时间。一个人浏览网页,网页上有n个标签,这个人需要L,R的区间的标签,其余区间的标签关闭掉,让我们求最少多少秒能够只剩下他需要的区间。每次鼠标移动一个位置花费一秒,关掉某个区间也要花费一秒。

题解:模拟 分类讨论           我们大体上分为三大类,pos鼠标在L左边,pos区间R右边,pos在区间中间,然后还要考虑,L,R和n的关系。
一般人的思路:我就是

#include<bits/stdc++.h> using namespace std; int main() {     int n,pos,l,r,ans;     while(cin>>n>>pos>>l>>r)     {         ans=0;         if(l==1&&r==n)             ans=0;         else if(pos<=l&&r==n)             ans+=l-pos+1;         else if(pos>l&&r==n)             ans+=pos-l+1;         else if(pos>=r&&l==1)             ans+=pos-r+1;         else if(pos<r&&l==1)             ans+=r-pos+1;         else             ans+=min(abs(l-pos)+r-l,abs(pos-r)+(r-l))+2;         cout<<ans<<endl;     }     return 0; }

大神,聪明的人的思路:
#include<bits/stdc++.h> int a,b,c,d; using namespace std; int main(){ cin>>a>>b>>c>>d; if(c==1 &&d==a) cout<<0; else if(c==1) cout<<abs(d-b)+1; else if(d==a) cout<<abs(c-b)+1; else cout<<min(abs(c-b)+(d-c),abs(d-b)+(d-c))+2; return 0; }

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