题目：
Alice and Bob are playing a game. The game involves splitting up game pieces into two teams. There are n pieces, and the i-th piece has a strength pi.
The way to split up game pieces is split into several steps:
- First, Alice will split the pieces into two different groups A and B. This can be seen as writing the assignment of teams of a piece in an n character string, where each character is A or B.
- Bob will then choose an arbitrary prefix or suffix of the string, and flip each character in that suffix (i.e. change A to B and B to A). He can do this step at most once.
- Alice will get all the pieces marked A and Bob will get all the pieces marked B.
The strength of a player is then the sum of strengths of the pieces in the group.
Given Alice's initial split into two teams, help Bob determine an optimal strategy. Return the maximum strength he can achieve.
Input
The first line contains integer n (1 ≤ n ≤ 5·105) — the number of game pieces.
The second line contains n integers pi (1 ≤ pi ≤ 109) — the strength of the i-th piece.
The third line contains n characters A or B — the assignment of teams after the first step (after Alice's step).
Output
Print the only integer a — the maximum strength Bob can achieve.
Examples
Input
5 1 2 3 4 5 ABABA
Output
11
Input
5 1 2 3 4 5 AAAAA
Output
15
Input
1 1 B
Output
1
Note
In the first sample Bob should flip the suffix of length one.
In the second sample Bob should flip the prefix or the suffix (here it is the same) of length 5.
In the third sample Bob should do nothing.
解题报告：就是问给定字符串AB组成的，每个位置代表一个数字，在有一次可以改变任意前缀和后缀和的机会下（使其AB颠倒），怎么能使B的求和最大。刚上来进入误区了，就是想使B尽量往后跑，没注意到这个不是升序的。
可以开辟a b两个数组，用来存放前缀和（a是全部的前缀和，b是仅仅b字符所对下标的前缀和
），之后在循环下，将单点的前缀更新和后缀更新取最大，即可。
ac代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn=5e5+1000;
ll num[maxn];
ll a[maxn],b[maxn];
int n;
char str[maxn];
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%lld",&num[i]);
cin>>str+1;
a[0]=b[0]=0;
for(int i=1;i<=n;i++)
{
if(str[i]=='B')
{
b[i]=b[i-1]+num[i];
}
else
b[i]=b[i-1];
a[i]=a[i-1]+num[i];
}
ll ans=0;
for(int i=1;i<=n;i++)
{
ans=max(ans,(a[i]-b[i]+b[n]-b[i]));//前缀和更新，将前i位反转
ans=max(ans,(a[n]-a[i]-(b[n]-b[i])+b[i]));//后缀和更新。将后i-n位反转
}
printf("%lld\n",ans);
}