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.
题意：只能改变一次，把A改成B或者把B改成A。求最后B所能达到的最大值。
想明白就挺简单的，先把原来B的数值加起来。然后从前到后遍历一遍，找一个最大值，从后往前遍历一次，找一个最大值。比较就好了。。
代码如下：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define ll long long
using namespace std;
const int maxx=5e5+10;
int a[maxx];
char s[maxx];
int n;
int main()
{
while(scanf("%d",&n)!=EOF)
{
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
scanf("%s",s);
ll ans=0;
for(int i=0;i<n;i++)
{
if(s[i]=='B')
ans+=a[i];
}
ll sum1=ans;
ll sum2=ans;
for(int i=0;i<n;i++)
{
if(s[i]=='A')
sum1+=a[i];
else sum1-=a[i];
ans=max(sum1,ans);
}
for(int i=n-1;i>=0;i--)
{
if(s[i]=='A')
sum2+=a[i];
else sum2-=a[i];
ans=max(ans,sum2);
}
printf("%lld\n",ans);
}
return 0;
}
努力加油a啊，(o)/~