Pushok the dog has been chasing Imp for a few hours already.

Fortunately, Imp knows that Pushok is afraid of a robot vacuum cleaner.

While moving, the robot generates a string t consisting of letters 's' and 'h', that produces a lot of noise. We define noise of string t as the number of occurrences of string "sh" as a subsequence in it, in other words, the number of such pairs (i, j), that i < j and  and .

The robot is off at the moment. Imp knows that it has a sequence of strings ti in its memory, and he can arbitrary change their order. When the robot is started, it generates the string t as a concatenation of these strings in the given order. The noise of the resulting string equals the noise of this concatenation.

Help Imp to find the maximum noise he can achieve by changing the order of the strings.

Input

The first line contains a single integer n (1 ≤ n ≤ 105) — the number of strings in robot's memory.

Next n lines contain the strings t1, t2, ..., tn, one per line. It is guaranteed that the strings are non-empty, contain only English letters 's' and 'h' and their total length does not exceed 105.

Output

Print a single integer — the maxumum possible noise Imp can achieve by changing the order of the strings.

Examples

Input
4
ssh
hs
s
hhhs
Output
18
Input
2
h
s
Output
1

题意:
给定n个字符串自由组合,求最多出现多少个sh这样的子序列。
思路:
对于s1,s2,产生字符串的个数就是S1.SH+s2*SH+max(s1.S*s2.H+s2.S*s1.H)
所以按照s1.S*s2.H-s2.S*s1.H排序即可。
#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>
#define fuck(x) cout<<#x<<" = "<<x<<endl;
#define debug(a,i) cout<<#a<<"["<<i<<"] = "<<a[i]<<endl;
#define ls (t<<1)
#define rs ((t<<1)+1)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = ;
const int maxm = ;
const int inf = 2.1e9;
const ll Inf = ;
const int mod = ;
const double eps = 1e-;
const double pi = acos(-); struct node{
ll S,H;
}a[maxn]; char s[maxn]; bool cmp(node a,node b){
return a.S*b.H>a.H*b.S;
} int main()
{
// ios::sync_with_stdio(false);
// freopen("in.txt","r",stdin); int n;
scanf("%d",&n);
ll ans=;
for(int i=;i<=n;i++){
scanf("%s",s);
for(int j=;s[j];j++){
if(s[j]=='s'){a[i].S++;}
else{
ans+=a[i].S;
a[i].H++;
}
}
} sort(a+,a++n,cmp); ll tmp=;
for(int i=;i<=n;i++){ ans+=a[i].H*tmp;
tmp+=a[i].S;
}
printf("%lld\n",ans); return ;
}

CodeForces - 922D Robot Vacuum Cleaner (贪心)的更多相关文章

  1. Codeforces 922 C - Robot Vacuum Cleaner (贪心、数据结构、sort中的cmp)

    题目链接:点击打开链接 Pushok the dog has been chasing Imp for a few hours already. Fortunately, Imp knows that ...

  2. CF922D Robot Vacuum Cleaner 贪心+排序

    正确的贪心方法:按照比例排序. code: #include <bits/stdc++.h> #define N 200000 #define ll long long #define s ...

  3. Codeforces Round #461 (Div. 2) D. Robot Vacuum Cleaner

    D. Robot Vacuum Cleaner time limit per test 1 second memory limit per test 256 megabytes Problem Des ...

  4. 【Codeforces 922D】Robot Vacuum Cleaner

    [链接] 我是链接,点我呀:) [题意] 让你把n个字符串重新排序,然后按顺序连接在一起 使得这个组成的字符串的"sh"子序列最多 [题解] /* * 假设A的情况好于B * 也就 ...

  5. codeforces Gym 100338E Numbers (贪心,实现)

    题目:http://codeforces.com/gym/100338/attachments 贪心,每次枚举10的i次幂,除k后取余数r在用k-r补在10的幂上作为候选答案. #include< ...

  6. [Codeforces 1214A]Optimal Currency Exchange(贪心)

    [Codeforces 1214A]Optimal Currency Exchange(贪心) 题面 题面较长,略 分析 这个A题稍微有点思维难度,比赛的时候被孙了一下 贪心的思路是,我们换面值越小的 ...

  7. Codeforces 583 DIV2 Robot's Task 贪心

    原题链接:http://codeforces.com/problemset/problem/583/B 题意: 就..要打开一个电脑,必须至少先打开其他若干电脑,每次转向有个花费,让你设计一个序列,使 ...

  8. 489. Robot Room Cleaner扫地机器人

    [抄题]: Given a robot cleaner in a room modeled as a grid. Each cell in the grid can be empty or block ...

  9. codeforces 349B Color the Fence 贪心,思维

    1.codeforces 349B    Color the Fence 2.链接:http://codeforces.com/problemset/problem/349/B 3.总结: 刷栅栏.1 ...

随机推荐

  1. Hdu 1867 KMP

    题目链接 题目意思: 给出两个字符串a, b, 求最长的公共字串c, c是a的后缀,也是b的前缀. 本题没有具体说明哪个字符串是文本串和匹配串, 所以都要考虑 思路: 查找的时候, 当文本串结束的时候 ...

  2. 使用pip出现 cannot import name "main"

    最近在linux使用pip install时遇到了这个报错 1.jpg ImportError: cannot import name main 遇到这个问题,我的解决办法是:cd 到usr/bin目 ...

  3. 2017 校赛 问题 E: 神奇的序列

    题目描述        Aurora在南宁发现了一个神奇的序列,即对于该序列的任意相邻两数之和都不是三的倍数.现在给你一个长度为n的整数序列,让你判断是否能够通过重新排列序列里的数字使得该序列变成一个 ...

  4. 2018-8-10-三种方式设置特定设备UWP-XAML-view

    title author date CreateTime categories 三种方式设置特定设备UWP XAML view lindexi 2018-08-10 19:16:52 +0800 20 ...

  5. 【NS2】各种TCP版本 之 TCP Tahoe 和 TCP Reno(转载)

    实验目的 学习TCP的拥塞控制机制,并了解TCP Tahoe 和 TCP Reno的运行方式. 基础知识回顾 TCP/IP (Transmission Control Protocol/Interne ...

  6. LeetCode108 Convert Sorted Array to Binary Search Tree

    Given an array where elements are sorted in ascending order, convert it to a height balanced BST. (M ...

  7. 括号序列问题 uva 1626 poj 1141【区间dp】

    首先考虑下面的问题:Code[VS] 3657 我们用以下规则定义一个合法的括号序列: (1)空序列是合法的 (2)假如S是一个合法的序列,则 (S) 和[S]都是合法的 (3)假如A 和 B 都是合 ...

  8. @atcoder - AGC040C@ Neither AB nor BA

    目录 @description@ @solution@ @accepted code@ @detail@ @description@ 给定偶数 N,求由 'A', 'B', 'C' 三种字符组成的字符 ...

  9. jQuery 链

    通过 jQuery,可以把动作/方法链接在一起. Chaining 允许我们在一条语句中运行多个 jQuery 方法(在相同的元素上). jQuery 方法链接 直到现在,我们都是一次写一条 jQue ...

  10. 深入Java线程管理(五):线程池

    这几天主要是狂看源程序,在弥补了一些以前知识空白的同时,也学会了不少新的知识(比如 NIO),或者称为新技术吧. 线程池就是其中之一,一提到线程,我们会想到以前<操作系统>的生产者与消费者 ...