CodeForces 632B Alice, Bob, Two Teams

 Alice, Bob, Two Teams

Time Limit:1500MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Description

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 p_i.

The way to split up game pieces is split into several steps:

1. 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.
2. 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.
3. 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 p_i (1 ≤ p_i ≤ 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.

Sample Input

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

题解：首先要注意要用__int64定义ans等，这个题就是要先求出原序列中 B的价值，然后从前往后遍历，遇到 A就变成 B，相应地更新 B的价值 ans. 遇到 B就变成 A，相应地更新 B的价值 ans.再从后往前遍历，进行上述相同操作，最后输出 ans的最大值。

<div style="text-align: left;"><pre name="code" class="cpp"><span style="font-family:SimSun;">#include<cstdio>#include<cstring>#include<algorithm>using namespace std;struct node{__int64 a;char c;}s[500010];int main(){int n;while(scanf("%d",&n)!=EOF){for(int i=0;i<n;i++)scanf("%I64d",&s[i].a);getchar();for(int i=0;i<n;i++)scanf("%c",&s[i].c);__int64 ans=0;for(int i=0;i<n;i++)if(s[i].c=='B')ans+=s[i].a;__int64 t=ans,t2=ans;for(int i=0;i<n;i++){if(s[i].c=='B')t-=s[i].a;elset+=s[i].a;ans=max(t,ans);}t=t2;//由于此时ans的值已经改变，所以才另设了这个变量t2for(int i=n-1;i>=0;i--){if(s[i].c=='B')t-=s[i].a;elset+=s[i].a;ans=max(t,ans);}printf("%I64d\n",ans);}return 0;}</span>