Browser CodeForces - 915B

BrowserCodeForces - 915BLuba is surfing the Internet. She currently has n opened tabs in her browser, indexed from1 to n from left to right. The mouse cursor is currently located at th

Guuuuuu老师儿

429人浏览 · 2018-01-23 10:09:39

Guuuuuu老师儿 · 2018-01-23 10:09:39 发布

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].

一共有四种操作 1.左移

2.右移

3.关闭左边所有网页

4.关闭右边所有网页

每个操作都花费1秒问关闭区间外的网页最短花费多长时间

code:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
int main(){
    int n,pos,l,r;
    scanf("%d%d%d%d",&n,&pos,&l,&r);
    if(l == 1 && r == n)//没有需要关闭的网页
        printf("0");
    else if(l == 1){
        int a = abs(r-pos);
        printf("%d",a+1);
    }//只需要关闭右边的网页，两步，移到右边，关闭右边所有网页

    else if(r == n){
        int a = abs(pos-l);
        printf("%d",a+1);
    }//只需要关闭左边的网页，两步，移到左边，关闭左边所有网页

    else{
        int a = abs(pos-l);
        int b = abs(r-pos);
        printf("%d",min(a,b)+2+(r-l));
    }

    //第一步移到所花费的时间最少的一边，删去所有网页，然后再去另一边，需要经过一次l-r再删去另一边
    return 0;
}

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