Educational Codeforces Round 9 -- B - Alice, Bob, Two Teams

B. Alice, Bob, Two Teams

time limit per test

1.5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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:

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 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.

大体题意：

输入n，代表有n个数，然后再输入n个力量，在输入长度为n的字符串（字符串只包括A或

B），Bob 能够选择该字符串的某一前缀或者后缀，或者什么也不做，将其反转（如果是

A，则变成B，如果是B 则变成A），最后Bob的力量是所有B对应的力量和。输出最大的力

量和！

思路：

因为是前缀或者后缀，所以直接扫描终点即可！然后区间内的把A的力量加起来，区间外的

把B的力量加起来，比较最大值即可！

技巧：

可以开两个数组，利用容斥定理，记录A的力量和，和B的力量和，这样扫描终点只需要相减

数组即可！

最后记得用long long 存啊！

代码如下：

#include
using namespace std;
const int maxn = 500000 + 100;
const int maxt = 100 + 10;
const int INF = 1e8;
const double eps = 1e-8;
const double pi = acos(-1.0);
typedef long long ll;
typedef unsigned long long llu;
int a[maxn];
char s[maxn];
ll suma[maxn];
ll sumb[maxn];
int main()
{
    int n;
    while(scanf("%d",&n) == 1){
        for (int i = 1; i <= n; ++i){
            scanf("%d",&a[i]);
        }
        scanf("%s",s+1);
        for (int i = 1; i <= n; ++i){
            suma[i]=sumb[i]=0;
            suma[i]+=suma[i-1];
            sumb[i]+=sumb[i-1];
            if (s[i] == 'A')suma[i]+=a[i];
            else sumb[i]+=a[i];
        }
        int len = n;
        ll sum = -1;
        ll tsum=0;
        for (int i = 1; i <= n; ++i){
            tsum = suma[i] + sumb[len]-sumb[i];
            sum = max(sum,tsum);
        }
        for (int i = len; i > 0; --i){
            tsum = suma[len]-suma[i] + sumb[i];
            sum = max(sum,tsum);
        }
        printf("%I64d\n",sum);
    }
    return 0;
}