Codeforces Round #549 (Div. 2)A. The Doors

最新推荐文章于 2024-08-28 22:49:31 发布

转载最新推荐文章于 2024-08-28 22:49:31 发布 · 175 阅读

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原文链接：http://www.cnblogs.com/yewanting/p/10727214.html

本文介绍了一个有趣的编程问题，即在一系列门（代表左或右出口）中找到最快的逃离路径。通过遍历序列并跟踪左右出口门的状态，可以确定最早能够从任一出口离开的时刻。

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Three years have passes and nothing changed. It is still raining in London, and Mr. Black has to close all the doors in his home in order to not be flooded. Once, however, Mr. Black became so nervous that he opened one door, then another, then one more and so on until he opened all the doors in his house.

There are exactly two exits from Mr. Black's house, let's name them left and right exits. There are several doors in each of the exits, so each door in Mr. Black's house is located either in the left or in the right exit. You know where each door is located. Initially all the doors are closed. Mr. Black can exit the house if and only if all doors in at least one of the exits is open. You are given a sequence in which Mr. Black opened the doors, please find the smallest index

We have to note that Mr. Black opened each door at most once, and in the end all doors became open.

Input

The first line contains integer

The next line contains

It is guaranteed that there is at least one door located in the left exit and there is at least one door located in the right exit.

Output

Print the smallest integer

Examples

Input

5
0 0 1 0 0

Output

Input

4
1 0 0 1

Output

Note

In the first example the first two doors are from the left exit, so when Mr. Black opened both of them only, there were two more closed door in the left exit and one closed door in the right exit. So Mr. Black wasn't able to exit at that moment.

When he opened the third door, all doors from the right exit became open, so Mr. Black was able to exit the house.

In the second example when the first two doors were opened, there was open closed door in each of the exit.

With three doors opened Mr. Black was able to use the left exit.

解题思路：这道题就是给你n个数，0代表左边的门，1代表右边的门，只有把左边或者右边全部的门全部打开才可逃离，问最快逃离的序号；

实际上我们只需遍历一遍，看是先把左边的门遍历完还是右边的门遍历完即可，再输出答案；

 1 #include<iostream>
 2 #include<string.h>
 3 using namespace std;
 4 
 5 int n ;
 6 int a[200005];
 7 int olen = 0;         //0的个数，左边门的个数；
 8 int ylen = 0;         //1的个数，右边门的个数；
 9 int tmpo = 0;       //用来遍历统计左边的门的个数；
10 int tmpy = 0;      //用来遍历统计左边的门的个数；
11 int main()
12 {
13     cin>>n;
14     for(int i = 1 ; i <= n ; i++ )
15     {
16         cin>>a[i];
17         if(a[i]==1)
18         {
19             ylen++;
20         }else
21         olen++;
22     }
23     for(int i = 1 ; i <= n ; i++)
24     {
25         if(a[i]==1)
26         {
27             tmpy++;
28         }else
29         if(a[i]==0)
30         tmpo++;
31         
32         if(tmpy==ylen)  //用来遍历统计左边的门的个数；如果左边先遍历完则输出；
33         {
34             cout<<i;
35             break;
36         }else
37         if(tmpo==olen)   //右边同理；
38         {
39             cout<<i;
40             break;
41         }
42     }
43     return 0;
44 }

代码如下：

转载于:https://www.cnblogs.com/yewanting/p/10727214.html