Codeforces Round #363 (Div. 2)->A. Launch of Collider

提示：扫一扫查出行
【扫一扫了解最新限行尾号】
复制提示

A. Launch of Collider

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There will be a launch of a new, powerful and unusual collider very soon, which located along a straight line. n particles will be launched inside it. All of them are located in a straight line and there can not be two or more particles located in the same point. The coordinates of the particles coincide with the distance in meters from the center of the collider, xi is the coordinate of the i-th particle and its position in the collider at the same time. All coordinates of particle positions are even integers.

You know the direction of each particle movement — it will move to the right or to the left after the collider's launch start. All particles begin to move simultaneously at the time of the collider's launch start. Each particle will move straight to the left or straight to the right with the constant speed of 1 meter per microsecond. The collider is big enough so particles can not leave it in the foreseeable time.

Write the program which finds the moment of the first collision of any two particles of the collider. In other words, find the number of microseconds before the first moment when any two particles are at the same point.

Input

The first line contains the positive integer n (1 ≤ n ≤ 200 000) — the number of particles.

The second line contains n symbols "L" and "R". If the i-th symbol equals "L", then the i-th particle will move to the left, otherwise the i-th symbol equals "R" and the i-th particle will move to the right.

The third line contains the sequence of pairwise distinct even integers x1, x2, ..., xn (0 ≤ xi ≤ 109) — the coordinates of particles in the order from the left to the right. It is guaranteed that the coordinates of particles are given in the increasing order.

Output

In the first line print the only integer — the first moment (in microseconds) when two particles are at the same point and there will be an explosion.

Print the only integer -1, if the collision of particles doesn't happen.

Examples

input

4
RLRL
2 4 6 10

output

input

3
LLR
40 50 60

output

-1

Note

In the first sample case the first explosion will happen in 1 microsecond because the particles number 1and 2 will simultaneously be at the same point with the coordinate 3.

In the second sample case there will be no explosion because there are no particles which will simultaneously be at the same point.

题意理解：有很多粒子在同一条直线上飞，速度是1，RL表示粒子的飞行方向，求粒子碰撞的最早时间，若不会碰撞就输出-1；

思路：只要是RL连着的就会碰撞，然后输出最小的minn/2;

#include<bits/stdc++.h>
using namespace std;
int a[];
int main() {
int n;
string s;
cin>>n>>s;
for(int i=; i<n; i++)
cin>>a[i];
int minn=-;
for(int i=; i<n-; i++) {
if(s[i]=='R'&&s[i+]=='L') {
if(minn==-)
minn=a[i+]-a[i];
else
minn=min(minn,a[i+]-a[i]);
}
}
if(minn!=-)
cout<<minn/<<endl;
else
cout<<minn<<endl;
return ;
}

Codeforces Round #363 (Div. 2)->A. Launch of Collider的更多相关文章

Codeforces Round 363 Div&period; 1 (A,B,C,D,E,F)
Codeforces Round 363 Div. 1 题目链接:## 点击打开链接 A. Vacations (1s, 256MB) 题目大意:给定连续 \(n\) 天,每天为如下四种状态之一: 不 ...
Codeforces Round &num;363 Div&period;2[111110]
好久没做手生了,不然前四道都是能A的,当然,正常发挥也是菜. A:Launch of Collider 题意:20万个点排在一条直线上,其坐标均为偶数.从某一时刻开始向左或向右运动,速度为每秒1个单位 ...
Codeforces Round &num;363 (Div&period; 2) A、B、C
A. Launch of Collider time limit per test 2 seconds memory limit per test 256 megabytes input standa ...
Codeforces Round &num;363 (Div&period; 2)
A题 http://codeforces.com/problemset/problem/699/A 非常的水,两个相向而行,且间距最小的点,搜一遍就是答案了. #include <cstdio& ...
Codeforces Round &num;363 (Div&period; 1) B&period; Fix a Tree 树的拆环
题目链接:http://codeforces.com/problemset/problem/698/B题意:告诉你n个节点当前的父节点,修改最少的点的父节点使之变成一棵有根树.思路:拆环.题解:htt ...
Codeforces Round &num;363 (Div&period; 2) A
Description There will be a launch of a new, powerful and unusual collider very soon, which located ...
Codeforces Round &num;363 (Div&period; 2) A 水
Description There will be a launch of a new, powerful and unusual collider very soon, which located ...
Codeforces Round &num;363 (Div&period; 2) D&period; Fix a Tree —— 并查集
题目链接:http://codeforces.com/contest/699/problem/D D. Fix a Tree time limit per test 2 seconds memory ...
Codeforces Round &num;363 (Div&period; 2) B&period; One Bomb —— 技巧
题目链接:http://codeforces.com/contest/699/problem/B 题解: 首先统计每行每列出现'*'的次数,以及'*'出现的总次数,得到r[n]和c[m]数组,以及su ...

随机推荐

【推荐】CentOS安装vsftpd-3&period;0&period;2+安全配置
注:以下所有操作均在CentOS 6.5 x86_64位系统下完成. FTP的登录一般有三种方式,分别是: 匿名用户形式:默认安装的情况下,系统只提供匿名用户访问,只需要输入用户anonymous/f ...
Web 开发人员系统重装备忘录
准备工作: 一.配置IIS 软件安装: 一.大块头: 1.VS2005 1.VS2005SP1 2.VSS 2005 2.VS2008 1.VS2008TeamExplorer 2.VS2008SP1 ...
25个增强iOS应用程序性能的提示和技巧--中级篇
25个增强iOS应用程序性能的提示和技巧--中级篇标签: ios性能优化内存管理 2013-12-13 10:55 738人阅读评论(0) 收藏举报分类: IPhone开发高级系列(34) ...
NSNumber 、 NSValue 、日期处理、集合类、 NSArray(一)
1 基本数据类型的封装 1.1 问题我们所学的所有基本数据类型,如int.float.double.char等,都不是对象,不能向它们发送消息.然而,在Foundation中的许多类,如NSArra ...
cin的返回值
例: int main() { int a,b; while(cin >> a >> b) cout << a+b << endl; } 首先,cin是 ...
sudo的使用和配置
1 sudo是什么 Sudo是Unix/Linux平台上的一个非常有用的工具,它允许系统管理员分配给普通用户一些合理的“权利”,让他们执行一些只有超级用户或其他特许用户才能完成的任务,比如:运行一些像 ...
jmeter负载机运行/添加压力机/分布式
• 我们在压测的时候,可能并发比较大, 一台机子已经启动不了那么多并发了,这个时候我们就要使用多台机子一起来发压力,就要添加压力机,添加压力机怎么添加呢,首先要在做压力机的机子上启动jmeter的代理 ...
修改 input[type=&quot&semi;radio&quot&semi;] 和 input[type=&quot&semi;checkbox&quot&semi;] 的默认样式
表单中,经常会使用到单选按钮和复选框,但是,input[type="radio"] 和 input[type="checkbox"] 的默认样式在不同的浏览器或 ...
Knockout&period;js Attr绑定
<head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8&quo ...
final关键字的特点
1.这个关键字是一个修饰符,可以修饰类,方法,变量. 2.被final修饰的类是一个最终类,不可以被继承. 3.被final修饰的方法是一个最终方法,不可以被覆盖. 4.被final修饰的变量是一个常 ...